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I'm trying to find a efficient solution for the next riddle:
i have a logical matrix at (n * n) size filled in false values
i need to create a function that will get zero or one as argument it will shift all
the values in the matrix one step to the left (meaning the first
element on the first row is deleted and the last element in the last
row is our new bit) and return true if there is a row/column in our
matrix contains only one's values.
No limitation on the data structure.
My naive solution in javascript:
const next = (bit, matrix) => {
matrix.shift()
matrix.push(bit);
const matrix_size = Math.sqrt(matrix.length);
let col_sum = 0;
let row_sum = 0;
for (let i = 0; i < matrix.length; ++i) {
col_sum = matrix[i];
row_sum += matrix[i];
if ((i + 1) % matrix_size === 0) {
if (row_sum === matrix_size) return true;
row_sum = 0;
}
for (let j = i + matrix_size;j < (i + ((matrix_size * matrix_size) - 1)); j += matrix_size) {
col_sum += matrix[j];
}
if (col_sum === matrix_size) return true;
}
return false;
}
i used 1d array as data structure but it doesn't really help my to reduce time complexity.
Love to hear some ideas :)
Let’s think about following example matrix:
[0, 0, 0, 0,
0, 0, 0, 0,
0, 0, 1, 1,
1, 1, 1, 1]
and push zero 16 times.
Then, False, True, True, True, False, True, True, True, False, True, True, True, False, False False and False will be obtained.
There is cyclic behavior (False, True, True, True).
If the length of continued ones was fixed, it isn’t necessary to recalculate every time in update.
Updated the matrix, the length of continued ones at top-left and bottom-right can be change, and it can be needed to update the cyclic memory.
Maintaining continued ones sequences, maintaining total count of cyclic behavior affected by the sequences, the complexity for the rows will be in O(1).
In case of column, instead of shifting and pushing, let matrix[cur]=bit and cur = (cur+1)%(matrix_size*matrix_size) to represent cur as the actual upper-left of the matrix.
Maintaining col_sum of each column, maintaining total count satisfying the all-ones-condition, the complexity will be O(1).
class Matrix:
def __init__(self, n):
self.mat = [0] * (n*n)
self.seq_len = [0] * (n*n)
self.col_total = [0] * n
self.col_archive = 0
self.row_cycle_cnt = [0] * n
self.cur = 0
self.continued_one = 0
self.n = n
def update(self, bit):
prev_bit = self.mat[self.cur]
self.mat[self.cur] = bit
# update col total
col = self.cur % self.n
if self.col_total[col] == self.n:
self.col_archive -= 1
self.col_total[col] += bit - prev_bit
if self.col_total[col] == self.n:
self.col_archive += 1
# update row index
# process shift out
if prev_bit == 1:
prev_len = self.seq_len[self.cur]
if prev_len > 1:
self.seq_len[(self.cur + 1) % (self.n * self.n)] = prev_len-1
if self.n <= prev_len and prev_len < self.n*2:
self.row_cycle_cnt[self.cur % self.n] -= 1
# process new bit
if bit == 0:
self.continued_one = 0
else:
self.continued_one = min(self.continued_one + 1, self.n*self.n)
# write the length of continued_one at the head of sequence
self.seq_len[self.cur+1 - self.continued_one] = self.continued_one
if self.n <= self.continued_one and self.continued_one < self.n*2:
self.row_cycle_cnt[(self.cur+1) % self.n] += 1
# update cursor
self.cur = (self.cur + 1) % (self.n * self.n)
return (self.col_archive > 0) or (self.row_cycle_cnt[self.cur % self.n] > 0)
def check2(self):
for y in range(self.n):
cnt = 0
for x in range(self.n):
cnt += self.mat[(self.cur + y*self.n + x) % (self.n*self.n)]
if cnt == self.n:
return True
for x in range(self.n):
cnt = 0
for y in range(self.n):
cnt += self.mat[(self.cur + y*self.n + x) % (self.n*self.n)]
if cnt == self.n:
return True
return False
if __name__ == "__main__":
import random
random.seed(123)
m = Matrix(4)
for i in range(100000):
ans1 = m.update(random.randint(0, 1))
ans2 = m.check2()
assert(ans1 == ans2)
print("epoch:{} mat={} ans={}".format(i, m.mat[m.cur:] + m.mat[:m.cur], ans1))
We are given an array with n values.
Example: [1,4,5,6,6]
For each index i of the array a ,we construct a new element of array b such that,
b[i]= [a[i]/1] + [a[i+1]/2] + [a[i+2]/3] + ⋯ + [a[n]/(n−i+1)] where [.] denotes the greatest integer function.
We are given an integer k as well.
We have to find the minimum i such that b[i] ≤ k.
I know the brute-force O(n^2) algorithm (to create the array - 'b'), can anybody suggest a better time complexity and way solve it?
For example, for the input [1,2,3],k=3, the output is 1(minimum-index).
Here, a[1]=1; a[2]=2; a[3]=3;
Now, b[1] = [a[1]/1] + [a[2]/2] + [a[3]/3] = [1/1] + [2/2] + [3/3] = 3;
b[2] = [a[2]/1] + [a[3]/2] = [2/1] + [3/2] = 3;
b[3] = [a[3]/1] = [3/1] = 3 (obvious)
Now, we have to find the index i such that b[i]<=k , k='3' , also b[1]<=3, henceforth, 1 is our answer! :-)
Constraints : - Time limits: -(2-seconds) , 1 <= a[i] <= 10^5, 1 <=
n <= 10^5, 1 <= k <= 10^9
Here's an O(n √A)-time algorithm to compute the b array where n is the number of elements in the a array and A is the maximum element of the a array.
This algorithm computes the difference sequence of the b array (∆b = b[0], b[1] - b[0], b[2] - b[1], ..., b[n-1] - b[n-2]) and derives b itself as the cumulative sums. Since the differences are linear, we can start with ∆b = 0, 0, ..., 0, loop over each element a[i], and add the difference sequence for [a[i]], [a[i]/2], [a[i]/3], ... at the appropriate spot. The key is that this difference sequence is sparse (less than 2√a[i] elements). For example, for a[i] = 36,
>>> [36//j for j in range(1,37)]
[36, 18, 12, 9, 7, 6, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
>>> list(map(operator.sub,_,[0]+_[:-1]))
[36, -18, -6, -3, -2, -1, -1, -1, 0, -1, 0, 0, -1, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
We can derive the difference sequence from a subroutine that, given a positive integer r, returns all maximal pairs of positive integers (p, q) such that pq ≤ r.
See complete Python code below.
def maximal_pairs(r):
p = 1
q = r
while p < q:
yield (p, q)
p += 1
q = r // p
while q > 0:
p = r // q
yield (p, q)
q -= 1
def compute_b_fast(a):
n = len(a)
delta_b = [0] * n
for i, ai in enumerate(a):
previous_j = i
for p, q in maximal_pairs(ai):
delta_b[previous_j] += q
j = i + p
if j >= n:
break
delta_b[j] -= q
previous_j = j
for i in range(1, n):
delta_b[i] += delta_b[i - 1]
return delta_b
def compute_b_slow(a):
n = len(a)
b = [0] * n
for i, ai in enumerate(a):
for j in range(n - i):
b[i + j] += ai // (j + 1)
return b
for n in range(1, 100):
print(list(maximal_pairs(n)))
lst = [1, 34, 3, 2, 9, 21, 3, 2, 2, 1]
print(compute_b_fast(lst))
print(compute_b_slow(lst))
This probably cannot reach the efficiency of David Eisenstat's answer but since I spent quite a long time figuring out an implementation, I thought I'd leave it up anyway. As it is, it seems about O(n^2).
The elements of b[i] may be out of order, but sections of them are not:
[a[1]/1] + [a[2]/2] + [a[3]/3]
|------ s2_1 -----|
|-s1_1-|
[a[2]/1] + [a[3]/2]
|------ s2_2 -----|
|-s1_2-|
[a[3]/1]
|-s1_3-|
s2_1 < s2_2
s1_1 < s1_2 < s1_3
Binary search for k on s1. Any result with an s1_i greater than k will rule out a section of ordered rows (rows are b_is).
Binary search for k on s2 on the remaining rows. Any result with an s2_i greater than k will rule out a section of ordered rows (rows are b_is).
This wouldn't help much since in the worst case, we'd have O(n^2 * log n) complexity, greater than O(n^2).
But we can also search horizontally. If we know that b_i ≤ k, then it will rule out both all rows with greater or equal length and the need to search smaller s(m)s, not because smaller s(m)s cannot produce a sum >= k, but because they will necessarily produce one with a higher i and we are looking for the minimum i.
JavaScript code:
var sum_width_iterations = 0
var total_width_summed = 0
var sum_width_cache = {}
function sum_width(A, i, width){
let key = `${i},${width}`
if (sum_width_cache.hasOwnProperty(key))
return sum_width_cache[key]
sum_width_iterations++
total_width_summed += width
let result = 0
for (let j=A.length-width; j<A.length; j++)
result += ~~(A[j] / (j + 1 - i))
return sum_width_cache[key] = result
}
function get_b(A){
let result = []
A.map(function(a, i){
result.push(sum_width(A, i, A.length - i))
})
return result
}
function find_s_greater_than_k(A, width, low, high, k){
let mid = low + ((high - low) >> 1)
let s = sum_width(A, mid, width)
while (low <= high){
mid = low + ((high - low) >> 1)
s = sum_width(A, mid, width)
if (s > k)
high = mid - 1
else
low = mid + 1
}
return [mid, s]
}
function f(A, k, l, r){
let n = A.length
if (l > r){
console.log(`l > r: l, r: ${l}, ${r}`)
return [n + 1, Infinity]
}
let width = n - l
console.log(`\n(call) width, l, r: ${width}, ${l}, ${r}`)
let mid = l + ((r - l) >> 1)
let mid_width = n - mid
console.log(`mid: ${mid}`)
console.log('mid_width: ' + mid_width)
let highest_i = n - mid_width
let [i, s] = find_s_greater_than_k(A, mid_width, 0, highest_i, k)
console.log(`hi_i, s,i,k: ${highest_i}, ${s}, ${i}, ${k}`)
if (mid_width == width)
return [i, s]
// either way we need to look left
// and down
console.log(`calling left`)
let [li, ls] = f(A, k, l, mid - 1)
// if i is the highest, width is
// the width of b_i
console.log(`got left: li, ls, i, high_i: ${li}, ${ls}, ${i}, ${highest_i}`)
if (i == highest_i){
console.log(`i == highest_i, s <= k: ${s <= k}`)
// b_i is small enough
if (s <= k){
if (ls <= k)
return [li, ls]
else
return [i, s]
// b_i is larger than k
} else {
console.log(`b_i > k`)
let [ri, rs] = f(A, k, mid + 1, r)
console.log(`ri, rs: ${ri}, ${rs}`)
if (ls <= k)
return [li, ls]
else if (rs <= k)
return [ri, rs]
else
return [i, s]
}
// i < highest_i
} else {
console.log(`i < highest_i: high_i, i, s, li, ls, mid, mid_width, width, l, r: ${highest_i}, ${i}, ${s}, ${li}, ${ls}, ${mid}, ${mid_width}, ${width}, ${l}, ${r}`)
// get the full sum for this b
let b_i = sum_width(A, i, n - i)
console.log(`b_i: ${b_i}`)
// suffix sum is less than k
// so we cannot rule out either side
if (s < k){
console.log(`s < k`)
let ll = l
let lr = mid - 1
let [lli, lls] = f(A, k, ll, lr)
console.log(`ll, lr, lli, lls: ${ll}, ${lr}, ${lli}, ${lls}`)
// b_i is a match so we don't
// need to look to the right
if (b_i <= k){
console.log(`b_i <= k: i, b_i: ${i}, ${b_i}`)
if (lls <= k)
return [lli, lls]
else
return [i, b_i]
// b_i > k
} else {
console.log(`b_i > k: i, b_i: ${i}, ${b_i}`)
let rl = mid + 1
let rr = r
let [rri, rrs] = f(A, k, rl, rr)
console.log(`rl, rr, rri, rrs: ${rl}, ${rr}, ${rri}, ${rrs}`)
// return the best of right
// and left sections
if (lls <= k)
return [lli, lls]
else if (rrs <= k)
return [rri, rrs]
else
return [i, b_i]
}
// suffix sum is greater than or
// equal to k so we can rule out
// this and all higher rows (`b`s)
// that share this suffix
} else {
console.log(`s >= k`)
let ll = l
// the suffix rules out b_i
// and above
let lr = i - 1
let [lli, lls] = f(A, k, ll, lr)
console.log(`ll, lr, lli, lls: ${ll}, ${lr}, ${lli}, ${lls}`)
let rl = highest_i + 1
let rr = r
let [rri, rrs] = f(A, k, rl, rr)
console.log(`rl, rr, rri, rrs: ${rl}, ${rr}, ${rri}, ${rrs}`)
// return the best of right
// and left sections
if (lls <= k)
return [lli, lls]
else if (rrs <= k)
return [rri, rrs]
else
return [i, b_i]
}
}
}
let lst = [1, 2, 3, 1]
// b [3, 3, 3, 1]
lst = [ 1, 34, 3, 2, 9, 21, 3, 2, 2, 1]
// b [23, 41, 12, 13, 20, 22, 4, 3, 2, 1]
console.log(
JSON.stringify(f(lst, 20, 0, lst.length)))
console.log(`sum_width_iterations: ${sum_width_iterations}`)
console.log(`total_width_summed: ${total_width_summed}`)
Why should calculating b[i] lead to O(n²)? If i = 1, it takes n steps. If i = n, it takes one step to calculate b[i]...
You could improve your calculation when you abort the sum on the condition Sum > k.
Let a in N^n
Let k in N
for (i1 := 1; i1 <= n; i1++)
b := 0
for (i2 :=i1; i2 <= n; i2++) // This loop is the calculation of b[i]
b := b + ceil(a[i2]/(i2 + 1))
if (b > k)
break
if (i2 == n)
return i1
I have problem with my homework.
Given a board of dimensions m x n is given, cut this board into rectangular pieces with the best total price. A matrix gives the price for each possible board size up through the original, uncut board.
Consider a 2 x 2 board with the price matrix:
3 4
3 6
We have a constant cost for each cutting for example 1.
Piece of length 1 x 1 is worth 3.
Horizontal piece of length 1 x 2 is worth 4.
Vertical piece of length 1 x 2 is worth 3.
Whole board is worth 6.
For this example, the optimal profit is 9, because we cut board into 1 x 1 pieces. Each piece is worth 3 and we done a 3 cut, so 4 x 3 - 3 x 1 = 9.
Second example:
1 2
3 4
Now I have to consider all the solutions:
4 1x1 pieces is worth 4x1 - (cost of cutting) 3x1 = 1
2 horizontal 1x2 is worth 2x2 - (cost of cutting) 1x1 = 3
2 vertical 1x2 is worth 3x2 - (cost of cutting) 1x1 = 5 -> best optimal profit
1 horizontal 1x2 + 2 x (1x1) pieces is worth 2 + 2 - (cost of cutting) 2 = 2
1 vertical 1x2 + 2 x (1x1) pieces is worth 3 + 2 - (cost of cutting) 2 = 3
I've read a lot about rod cutting algorithm but I don't have any idea how to bite this problem.
Do you have any ideas?
I did this in Python. The algorithm is
best_val = value of current board
check each horizontal and vertical cut for better value
for cut point <= half the current dimension (if none, return initial value)
recur on the two boards formed
if sum of values > best_val
... best_val = that sum
... record cut point and direction
return result: best_val, cut point, and direction
I'm not sure what you'll want for return values; I gave back the best value and tree of boards. For your second example, this is
(5, [[2, 1], [2, 1]])
Code, with debugging traces (indent and the labeled prints):
indent = ""
indent_len = 3
value = [[1, 2],
[3, 4]]
def best_cut(high, wide):
global indent
print indent, "ENTER", high, wide
indent += " " * indent_len
best_val = value[high-1][wide-1]
print indent, "Default", best_val
cut_vert = None
cut_val = best_val
cut_list = []
# Check horizontal cuts
for h_cut in range(1, 1 + high // 2):
print indent, "H_CUT", h_cut
cut_val1, cut_list1 = best_cut(h_cut, wide)
cut_val2, cut_list2 = best_cut(high - h_cut, wide)
cut_val = cut_val1 + cut_val2
if cut_val > best_val:
cut_list = [cut_list1, cut_list2]
print indent, "NEW H", h_cut, cut_val, cut_list
best_val = cut_val
cut_vert = False
best_h = h_cut
# Check vertical cuts
for v_cut in range(1, 1 + wide // 2):
print indent, "V_CUT", v_cut
cut_val1, cut_list1 = best_cut(high, v_cut)
cut_val2, cut_list2 = best_cut(high, wide - v_cut)
cut_val = cut_val1 + cut_val2
if cut_val > best_val:
cut_list = [cut_list1, cut_list2]
print indent, "NEW V", v_cut, cut_val, cut_list
best_val = cut_val
cut_vert = True
best_v = v_cut
# Return result of best cut
# Remember to subtract the cut cost
if cut_vert is None:
result = best_val , [high, wide]
elif cut_vert:
result = best_val-1, cut_list
else:
result = best_val-1, cut_list
indent = indent[indent_len:]
print indent, "LEAVE", cut_vert, result
return result
print best_cut(2, 2)
Output (profit and cut sizes) for each of the two tests:
(9, [[[1, 1], [1, 1]], [[1, 1], [1, 1]]])
(5, [[2, 1], [2, 1]])
Let f(h,w) represent the best total price achievable for a board with height h and width w with cutting price c. Then
f(h,w) = max(
price_matrix(h, w),
f(i, w) + f(h - i, w) - c,
f(h, j) + f(h, w - j) - c
)
for i = 1 to floor(h / 2)
for j = 1 to floor(w / 2)
Here's a bottom-up example in JavaScript that returns the filled table given the price matrix. The answer would be in the bottom right corner.
function f(prices, cost){
var m = new Array(prices.length);
for (let i=0; i<prices.length; i++)
m[i] = [];
for (let h=0; h<prices.length; h++){
for (let w=0; w<prices[0].length; w++){
m[h][w] = prices[h][w];
if (h == 0 && w == 0)
continue;
for (let i=1; i<(h+1>>1)+1; i++)
m[h][w] = Math.max(
m[h][w],
m[i-1][w] + m[h-i][w] - cost
);
for (let i=1; i<(w+1>>1)+1; i++)
m[h][w] = Math.max(
m[h][w],
m[h][i-1] + m[h][w-i] - cost
);
}
}
return m;
}
$('#submit').click(function(){
let prices = JSON.parse($('#input').val());
let result = f(prices, 1);
let str = result.map(line => JSON.stringify(line)).join('<br>');
$('#output').html(str);
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<textarea id="input">[[3, 4],
[3, 6]]</textarea>
<p><button type="button" id="submit">Submit</button></p>
<div id="output"><div>
Some thoughts on the problem rather than an answer:
It was a long time ago i studied dynamic programming, but i wrote up the following pseudo code which is think is O(n^2):
// 'Board'-class not included
val valueOfBoards: HashMap<Board, int>
fun cutBoard(b: Board, value: int) : int {
if (b.isEmpty()) return 0
if (valueOfBoards[b] > value) {
return 0;
} else {
valueOfBoards[b] = value
}
int maxValue = Integer.MIN_VALUE
for (Board piece : b.getPossiblePieces()) {
val (cuttingCost, smallerBoard) = b.cutOffPiece(piece)
val valueGained: int = piece.getPrice() - cuttingCost
maxValue = Max(maxValue, valueGained + cutBoard(smallerBoard, value + valueGained))
}
return maxValue;
}
The board class is not trivially implemented, here is some elaboration:
// returns all boards which fits in the current board
// for the initial board this will be width*height subboards
board.getPossiblePieces()
// returns a smaller board and the cutting cost of the cut
// I can see this becoming complex, depends on how one chooses to represent the board.
board.cutOffPiece(piece: Board)
It is not clear to me at the moment if cutOffPiece() breaks the algorithm in that you do not know how to optimally cut. I think since the algorithm will proceed from larger pieces to smaller pieces at some point it will be fine.
I tried to solve the re computation of sub problems (identical boards) by storing results in something like HashMap<Board, price> and comparing the new board with the stored best price before proceeding.
According to your answers I've prepared bottom-up and top-down implementation.
Bottom-up:
function bottomUp($high, $wide, $matrix){
$m = [];
for($h = 0; $h < $high; $h++){
for($w = 0; $w < $wide; $w++){
$m[$h][$w] = $matrix[$h][$w];
if($h == 0 && $w == 0){
continue;
}
for($i = 1; $i < ($h + 1 >> 1) + 1; $i++){
$m[$h][$w] = max(
$m[$h][$w],
$m[$i - 1][$w] + $m[$h - $i][$w] - CUT_COST
);
}
for($i = 1; $i < ($w + 1 >> 1) + 1; $i++){
$m[$h][$w] = max(
$m[$h][$w],
$m[$h][$i - 1] + $m[$h][$w - $i] - CUT_COST
);
}
}
}
return $m[$high-1][$wide-1];
}
Top-down:
function getBestCut($high, $wide, $matrix){
global $checked;
if(isset($checked[$high][$wide])){
return $checked[$high][$wide];
}
$bestVal = $matrix[$high-1][$wide-1];
$cutVert = CUT_VERT_NONE;
$cutVal = $bestVal;
$cutList = [];
for($hCut = 1; $hCut < 1 + floor($high/2); $hCut++){
$result1 = getBestCut($hCut, $wide, $matrix);
$cutVal1 = $result1[0];
$cutList1 = $result1[1];
$result2 = getBestCut($high - $hCut, $wide, $matrix);
$cutVal2 = $result2[0];
$cutList2 = $result2[1];
$cutVal = $cutVal1 + $cutVal2;
if($cutVal > $bestVal){
$cutList = [$cutList1, $cutList2];
$bestVal = $cutVal;
$cutVert = CUT_VERT_FALSE;
$bestH = $hCut;
}
$checked[$hCut][$wide] = $result1;
$checked[$high - $hCut][$wide] = $result2;
}
for($vCut = 1; $vCut < 1 + floor($wide/2); $vCut++){
$result1 = getBestCut($hCut, $vCut, $matrix);
$cutVal1 = $result1[0];
$cutList1 = $result1[1];
$result2 = getBestCut($high, $wide - $vCut, $matrix);
$cutVal2 = $result2[0];
$cutList2 = $result2[1];
$cutVal = $cutVal1 + $cutVal2;
if($cutVal > $bestVal){
$cutList = [$cutList1, $cutList2];
$bestVal = $cutVal;
$cutVert = CUT_VERT_TRUE;
$bestH = $vCut;
}
$checked[$hCut][$vCut] = $result1;
$checked[$high][$wide - $vCut] = $result2;
}
if($cutVert == CUT_VERT_NONE){
$result = [$bestVal, [$high, $wide]];
}else if($cutVert == CUT_VERT_TRUE){
$result = [$bestVal - CUT_COST, $cutList];
}else{
$result = [$bestVal - CUT_COST, $cutList];
}
return $result;
}
Please tell me are they correct implementation of this method?
I wonder if time complexity is O(m^2*n^2) in top-down method?
A question last week defined the zig zag ordering on an n by m matrix and asked how to list the elements in that order.
My question is how to quickly find the ith item in the zigzag ordering? That is, without traversing the matrix (for large n and m that's much too slow).
For example with n=m=8 as in the picture and (x, y) describing (row, column)
f(0) = (0, 0)
f(1) = (0, 1)
f(2) = (1, 0)
f(3) = (2, 0)
f(4) = (1, 1)
...
f(63) = (7, 7)
Specific question: what is the ten billionth (1e10) item in the zigzag ordering of a million by million matrix?
Let's assume that the desired element is located in the upper half of the matrix. The length of the diagonals are 1, 2, 3 ..., n.
Let's find the desired diagonal. It satisfies the following property:
sum(1, 2 ..., k) >= pos but sum(1, 2, ..., k - 1) < pos. The sum of 1, 2, ..., k is k * (k + 1) / 2. So we just need to find the smallest integer k such that k * (k + 1) / 2 >= pos. We can either use a binary search or solve this quadratic inequality explicitly.
When we know the k, we just need to find the pos - (k - 1) * k / 2 element of this diagonal. We know where it starts and where we should move(up or down, depending on the parity of k), so we can find the desired cell using a simple formula.
This solution has an O(1) or an O(log n) time complexity(it depends on whether we use a binary search or solve the inequation explicitly in step 2).
If the desired element is located in the lower half of the matrix, we can solve this problem for a pos' = n * n - pos + 1 and then use symmetry to get the solution to the original problem.
I used 1-based indexing in this solution, using 0-based indexing might require adding +1 or -1 somewhere, but the idea of the solution is the same.
If the matrix is rectangular, not square, we need to consider the fact the length of diagonals look this way: 1, 2, 3, ..., m, m, m, .., m, m - 1, ..., 1(if m <= n) when we search for the k, so the sum becomes something like k * (k + 1) / 2 if k <= m and k * (k + 1) / 2 + m * (k - m) otherwise.
import math, random
def naive(n, m, ord, swap = False):
dx = 1
dy = -1
if swap:
dx, dy = dy, dx
cur = [0, 0]
for i in range(ord):
cur[0] += dy
cur[1] += dx
if cur[0] < 0 or cur[1] < 0 or cur[0] >= n or cur[1] >= m:
dx, dy = dy, dx
if cur[0] >= n:
cur[0] = n - 1
cur[1] += 2
if cur[1] >= m:
cur[1] = m - 1
cur[0] += 2
if cur[0] < 0: cur[0] = 0
if cur[1] < 0: cur[1] = 0
return cur
def fast(n, m, ord, swap = False):
if n < m:
x, y = fast(m, n, ord, not swap)
return [y, x]
alt = n * m - ord - 1
if alt < ord:
x, y = fast(n, m, alt, swap if (n + m) % 2 == 0 else not swap)
return [n - x - 1, m - y - 1]
if ord < (m * (m + 1) / 2):
diag = int((-1 + math.sqrt(1 + 8 * ord)) / 2)
parity = (diag + (0 if swap else 1)) % 2
within = ord - (diag * (diag + 1) / 2)
if parity: return [diag - within, within]
else: return [within, diag - within]
else:
ord -= (m * (m + 1) / 2)
diag = int(ord / m)
within = ord - diag * m
diag += m
parity = (diag + (0 if swap else 1)) % 2
if not parity:
within = m - within - 1
return [diag - within, within]
if __name__ == "__main__":
for i in range(1000):
n = random.randint(3, 100)
m = random.randint(3, 100)
ord = random.randint(0, n * m - 1)
swap = random.randint(0, 99) < 50
na = naive(n, m, ord, swap)
fa = fast(n, m, ord, swap)
assert na == fa, "(%d, %d, %d, %s) ==> (%s), (%s)" % (n, m, ord, swap, na, fa)
print fast(1000000, 1000000, 9999999999, False)
print fast(1000000, 1000000, 10000000000, False)
So the 10-billionth element (the one with ordinal 9999999999), and the 10-billion-first element (the one with ordinal 10^10) are:
[20331, 121089]
[20330, 121090]
An analytical solution
In the general case, your matrix will be divided in 3 areas:
an initial triangle t1
a skewed part mid where diagonals have a constant length
a final triangle t2
Let's call p the index of your diagonal run.
We want to define two functions x(p) and y(p) that give you the column and row of the pth cell.
Initial triangle
Let's look at the initial triangular part t1, where each new diagonal is one unit longer than the preceding.
Now let's call d the index of the diagonal that holds the cell, and
Sp = sum(di) for i in [0..p-1]
We have p = Sp + k, with 0 <=k <= d and
Sp = d(d+1)/2
if we solve for d, it brings
d²+d-2p = 0, a quadratic equation where we retain only the positive root:
d = (-1+sqrt(1+8*p))/2
Now we want the highest integer value closest to d, which is floor(d).
In the end, we have
p = d + k with d = floor((-1+sqrt(1+8*p))/2) and k = p - d(d+1)/2
Let's call
o(d) the function that equals 1 if d is odd and 0 otherwise, and
e(d) the function that equals 1 if d is even and 0 otherwise.
We can compute x(p) and y(p) like so:
d = floor((-1+sqrt(1+8*p))/2)
k = p - d(d+1)/2
o = d % 2
e = 1 - o
x = e*d + (o-e)*k
y = o*d + (e-o)*k
even and odd functions are used to try to salvage some clarity, but you can replace
e(p) with 1 - o(p) and have slightly more efficient but less symetric formulaes for x and y.
Middle part
let's consider the smallest matrix dimension s, i.e. s = min (m,n).
The previous formulaes hold until x or y (whichever comes first) reaches the value s.
The upper bound of p such as x(i) <= s and y(i) <= s for all i in [0..p]
(i.e. the cell indexed by p is inside the initial triangle t1) is given by
pt1 = s(s+1)/2.
For p >= pt1, diagonal length remains equal to s until we reach the second triangle t2.
when inside mid, we have:
p = s(s+1)/2 + ds + k with k in [0..s[.
which yields:
d = floor ((p - s(s+1)/2)/s)
k = p - ds
We can then use the same even/odd trick to compute x(p) and y(p):
p -= s(s+1)/2
d = floor (p / s)
k = p - d*s
o = (d+s) % 2
e = 1 - o
x = o*s + (e-o)*k
y = e*s + (o-e)*k
if (n > m)
x += d+e
y -= e
else
y += d+o
x -= o
Final triangle
Using symetry, we can calculate pt2 = m*n - s(s+1)/2
We now face nearly the same problem as for t1, except that the diagonal may run in the same direction as for t1 or in the reverse direction (if n+m is odd).
Using symetry tricks, we can compute x(p) and y(p) like so:
p = n*m -1 - p
d = floor((-1+sqrt(1+8*p))/2)
k = p - d*(d+1)/2
o = (d+m+n) % 2
e = 1 - $o;
x = n-1 - (o*d + (e-o)*k)
y = m-1 - (e*d + (o-e)*k)
Putting all together
Here is a sample c++ implementation.
I used 64 bits integers out of sheer lazyness. Most could be replaced by 32 bits values.
The computations could be made more effective by precomputing a few more coefficients.
A good part of the code could be factorized, but I doubt it is worth the effort.
Since this is just a quick and dirty proof of concept, I did not optimize it.
#include <cstdio> // printf
#include <algorithm> // min
using namespace std;
typedef long long tCoord;
void panic(const char * msg)
{
printf("PANIC: %s\n", msg);
exit(-1);
}
struct tPoint {
tCoord x, y;
tPoint(tCoord x = 0, tCoord y = 0) : x(x), y(y) {}
tPoint operator+(const tPoint & p) const { return{ x + p.x, y + p.y }; }
bool operator!=(const tPoint & p) const { return x != p.x || y != p.y; }
};
class tMatrix {
tCoord n, m; // dimensions
tCoord s; // smallest dimension
tCoord pt1, pt2; // t1 / mid / t2 limits for p
public:
tMatrix(tCoord n, tCoord m) : n(n), m(m)
{
s = min(n, m);
pt1 = (s*(s + 1)) / 2;
pt2 = n*m - pt1;
}
tPoint diagonal_cell(tCoord p)
{
tCoord x, y;
if (p < pt1) // inside t1
{
tCoord d = (tCoord)floor((-1 + sqrt(1 + 8 * p)) / 2);
tCoord k = p - (d*(d + 1)) / 2;
tCoord o = d % 2;
tCoord e = 1 - o;
x = o*d + (e - o)*k;
y = e*d + (o - e)*k;
}
else if (p < pt2) // inside mid
{
p -= pt1;
tCoord d = (tCoord)floor(p / s);
tCoord k = p - d*s;
tCoord o = (d + s) % 2;
tCoord e = 1 - o;
x = o*s + (e - o)*k;
y = e*s + (o - e)*k;
if (m > n) // vertical matrix
{
x -= o;
y += d + o;
}
else // horizontal matrix
{
x += d + e;
y -= e;
}
}
else // inside t2
{
p = n * m - 1 - p;
tCoord d = (tCoord)floor((-1 + sqrt(1 + 8 * p)) / 2);
tCoord k = p - (d*(d + 1)) / 2;
tCoord o = (d + m + n) % 2;
tCoord e = 1 - o;
x = n - 1 - (o*d + (e - o)*k);
y = m - 1 - (e*d + (o - e)*k);
}
return{ x, y };
}
void check(void)
{
tPoint move[4] = { { 1, 0 }, { -1, 1 }, { 1, -1 }, { 0, 1 } };
tPoint pos;
tCoord dir = 0;
for (tCoord p = 0; p != n * m ; p++)
{
tPoint dc = diagonal_cell(p);
if (pos != dc) panic("zot!");
pos = pos + move[dir];
if (dir == 0)
{
if (pos.y == m - 1) dir = 2;
else dir = 1;
}
else if (dir == 3)
{
if (pos.x == n - 1) dir = 1;
else dir = 2;
}
else if (dir == 1)
{
if (pos.y == m - 1) dir = 0;
else if (pos.x == 0) dir = 3;
}
else
{
if (pos.x == n - 1) dir = 3;
else if (pos.y == 0) dir = 0;
}
}
}
};
void main(void)
{
const tPoint dim[] = { { 10, 10 }, { 11, 11 }, { 10, 30 }, { 30, 10 }, { 10, 31 }, { 31, 10 }, { 11, 31 }, { 31, 11 } };
for (tPoint d : dim)
{
printf("Checking a %lldx%lld matrix...", d.x, d.y);
tMatrix(d.x, d.y).check();
printf("done\n");
}
tCoord p = 10000000000;
tMatrix matrix(1000000, 1000000);
tPoint cell = matrix.diagonal_cell(p);
printf("Coordinates of %lldth cell: (%lld,%lld)\n", p, cell.x, cell.y);
}
Results are checked against "manual" sweep of the matrix.
This "manual" sweep is a ugly hack that won't work for a one-row or one-column matrix, though diagonal_cell() does work on any matrix (the "diagonal" sweep becomes linear in that case).
The coordinates found for the 10.000.000.000th cell of a 1.000.000x1.000.000 matrix seem consistent, since the diagonal d on which the cell stands is about sqrt(2*1e10), approx. 141421, and the sum of cell coordinates is about equal to d (121090+20330 = 141420). Besides, it is also what the two other posters report.
I would say there is a good chance this lump of obfuscated code actually produces an O(1) solution to your problem.
I'm would like to create an algorithm that will divide text into 3-evenly sized groups (based on text length). Since this will be put to use for line-breaks, the order of the text needs to be maintained.
For instance this string:
Just testing to see how this works.
would sort to:
Just testing // 12 characters
to see how // 10 characters
this works. // 11 characters
Any ideas?
The "minimum raggedness" dynamic program, also from the Wikipedia article on word wrap, can be adapted to your needs. Set LineWidth = len(text)/n - 1 and ignore the comment about infinite penalties for exceeding the line width; use the definition of c(i, j) as is with P = 2.
Code. I took the liberty of modifying the DP always to return exactly n lines, at the cost of increasing the running time from O(#words ** 2) to O(#words ** 2 * n).
def minragged(text, n=3):
"""
>>> minragged('Just testing to see how this works.')
['Just testing', 'to see how', 'this works.']
>>> minragged('Just testing to see how this works.', 10)
['', '', 'Just', 'testing', 'to', 'see', 'how', 'this', 'works.', '']
"""
words = text.split()
cumwordwidth = [0]
# cumwordwidth[-1] is the last element
for word in words:
cumwordwidth.append(cumwordwidth[-1] + len(word))
totalwidth = cumwordwidth[-1] + len(words) - 1 # len(words) - 1 spaces
linewidth = float(totalwidth - (n - 1)) / float(n) # n - 1 line breaks
def cost(i, j):
"""
cost of a line words[i], ..., words[j - 1] (words[i:j])
"""
actuallinewidth = max(j - i - 1, 0) + (cumwordwidth[j] - cumwordwidth[i])
return (linewidth - float(actuallinewidth)) ** 2
# best[l][k][0] is the min total cost for words 0, ..., k - 1 on l lines
# best[l][k][1] is a minimizing index for the start of the last line
best = [[(0.0, None)] + [(float('inf'), None)] * len(words)]
# xrange(upper) is the interval 0, 1, ..., upper - 1
for l in xrange(1, n + 1):
best.append([])
for j in xrange(len(words) + 1):
best[l].append(min((best[l - 1][k][0] + cost(k, j), k) for k in xrange(j + 1)))
lines = []
b = len(words)
# xrange(upper, 0, -1) is the interval upper, upper - 1, ..., 1
for l in xrange(n, 0, -1):
a = best[l][b][1]
lines.append(' '.join(words[a:b]))
b = a
lines.reverse()
return lines
if __name__ == '__main__':
import doctest
doctest.testmod()
You can try the next simple heuristic for starters: Place to iterators in n/3 and 2n/3 and search for the closest space near each of them.
From http://en.wikipedia.org/wiki/Word_wrap:
SpaceLeft := LineWidth
for each Word in Text
if Width(Word) > SpaceLeft
insert line break before Word in Text
SpaceLeft := LineWidth - Width(Word)
else
SpaceLeft := SpaceLeft - (Width(Word) + SpaceWidth)
This method is used by many modern word processors, such as OpenOffice.org Writer and Microsoft Word. This algorithm is optimal in that it always puts the text on the minimum number of lines.
The answer from "someone" works fine. However, I had problems translating this into SWIFT code. Here is my translation for all those that are interested.
import Foundation
class SplitText{
typealias MinRag = (Float, Int) // meaning (cost for line (so far), word index)
// from http://stackoverflow.com/questions/6426017/word-wrap-to-x-lines-instead-of-maximum-width-least-raggedness?lq=1
class func splitText(text:String, numberOfLines:Int)-> [String]{
//preparations
var words = split(text, maxSplit:100, allowEmptySlices: false, isSeparator:{(s:Character)-> Bool in return s == " " || s == "\n"})
var cumwordwidth = [Int](); //cummulative word widths
cumwordwidth.append(0);
for word in words{
cumwordwidth.append(cumwordwidth[cumwordwidth.count - 1] + count(word));
}
var totalwidth = cumwordwidth[cumwordwidth.count - 1] + count(words) - 1;
var linewidth:Float = Float(totalwidth - (numberOfLines - 1)) / Float(numberOfLines)
// cost function for one line for words i .. j
var cost = { (i:Int,j:Int)-> Float in
var actuallinewidth = max(j - i - 1, 0) + (cumwordwidth[j] - cumwordwidth[i]);
var remainingWidth: Float = linewidth - Float(actuallinewidth)
return remainingWidth * remainingWidth
}
var best = [[MinRag]]()
var tmp = [MinRag]();
//ensure that data structure is initialised in a way that we start with adding the first word
tmp.append((0, -1));
for word in words {
tmp.append((Float.infinity , -1));
}
best.append(tmp);
//now we can start. We simply calculate the cost for all possible lines
for l in 1...numberOfLines {
tmp = [MinRag]()
for j in 0...words.count {
var min:MinRag = (best[l - 1][0].0 + cost(0, j), 0);
var k: Int
for k = 0; k < j + 1 ; ++k {
var loc:Float = best[l - 1][k].0 + cost(k, j);
if (loc < min.0 || (loc == min.0 && k < min.1)) {
min=(loc, k);
}
println("l=\(l), j=\(j), k=\(k), min=\(min)")
}
tmp.append(min);
}
best.append(tmp);
}
//now build the answer based on above calculations
var lines = [String]();
var b = words.count;
var o:Int
for o = numberOfLines; o > 0 ; --o {
var a = best[o][b].1;
lines.append(" ".join(words[a...b-1]));
b = a;
}
return reverse(lines);
}
}