1. Honors Track: Competitive Programming & Problem Solving Binary Indexed Trees
Christiaan Dirkx

2. Contents Problem Introduction Formalization Naive Implementation
Binary Indexed Trees
Binary Intermezzo
Operations
Extenstions
Problems

3. Why Binary Indexed Trees?
Consider the following problem:
A list of values A 1 3 -1 4
Sum of all elements in A?
Sum of the first 3 elements in A?
Sum of element 2 through 7?

4. Formalisation
Abstact Data Type We want to store a collection C of elements, each with an associated index (integer value). Operations Get(S, i): gets the value of element in S with index i Sum(S, i): gets the sum of all elements in S with index ≤ i Update(S, i, v): updates the value of element in S with index i to value v

5. Formalisation
Abstact Data Type We want to store a collection C of elements, each with an associated index (integer value). Operations Get(S, i): gets the value of element in S with index i Sum(S, i): gets the sum of all elements in S with index ≤ i Update(S, i, v): updates the value of element in S with index i to value v Sum(S, a, b)?: gets the sum of all elements in S with index i, a ≤ i ≤ b

6. Formalisation
Abstact Data Type We want to store a collection C of elements, each with an associated index (integer value). Operations Get(S, i): gets the value of element in S with index i Sum(S, i): gets the sum of all elements in S with index ≤ i Update(S, i, v): updates the value of element in S with index i to value v Sum(S, a, b): Sum(S, b) – Sum(S, a - 1)

7. Implementation
How can we implement this ADT?
Heaps!
(Not quite)

8. Implementation A: 1 3 -1 4 1 4 3 3 7 8 9
Aggregation array

9. Implementation A: Can we improve? Aggregation array 1 3 -1 4 1 4 3 3 71 3 -1 4 1 4 3 3 7 8 9
Aggregation array
Complexity?
Get(S, i):
O(1)
Can we improve?
Sum(S, i):
O(1)
Update(S, i, v):
O(n)

10. Implementation A: Can we improve? Aggregation array 1 3 -1 4 1 4 3 3 71 3 -1 4 1 4 3 3 7 8 9
Aggregation array
Complexity?
Get(S, i):
O(log n)
Can we improve?
Sum(S, i):
O(log n)
Update(S, i):
O(log n)

11. Binary Indexed Tree ? Change the recurrence relation The key idea:
Aggregation array:
Sum(S, 0): Get(S, 0)
Sum(S, i): Sum(S, i-1, i) + Sum(S, i-1)

12. Binary Indexed Tree ? ? Change the recurrence relation The key idea:
Aggregation array:
Sum(S, 0): Get(S, 0)
Sum(S, i): Get(S, i) + Sum(S, i-1) ?
Binary indexed tree:
O(log n)

13. Binary recurrence Sum(S, 13): Sum(S, 12, 13) + Sum(S, 12)
Consider the number 13
Sum(S, 13): Sum(S, 12, 13) + Sum(S, 12)
Sum(S, 12): Sum(S, 8, 12) + Sum(S, 8)
Sum(S, 8): Sum(S, 0, 8) + Sum(S, 0)
Why?
In binary:
8 1000

14. Binary recurrence Sum(S, 1101): Sum(S, 1100, 1101) + Sum(S, 1100)
Consider the number 13
Sum(S, 1101): Sum(S, 1100, 1101) + Sum(S, 1100)
Sum(S, 1100): Sum(S, 1000, 1100) + Sum(S, 1000)
Sum(S, 1000): Sum(S, 0000, 1000) + Sum(S, 0000)
O(log n)

15. Visualization
0000
0001
0010
0100
1000
0011
0101
0110
0111 A: 1 3 -1 4

16. Visualization 1 3 9 3 4 1 A: 1 3 -1 4 1 4 3 3 7 8 9

17. Visualization 1 3 9 3 4 1 A: 1 3 -1 4 1 4 3 3 7 8 9

18. Visualization T: A: But how should we encode the index? 1 3 3 4 1 9 11 3 3 4 1 9
But how should we encode the index? A: 1 3 -1 4 1 4 3 3 7 8 9

19. We want to iteratively get the least significant bit
Binary Intermezzo
Binary number:
most
significant bit
least
significant bit
We want to iteratively get the least significant bit

20. Binary Intermezzo i : 0 1 0 0 1 1 0 0 ~i : 1 0 1 1 0 0 1 1 -i :
Next index:
i :
~i :
-i :
i & -i :
i - (i & -i) :

21. Sum(S, i) A: T: Sum(S, i) Sum(S, 7) 1 3 -1 4 1 1 1 3 3 4 1 9 sum: 01 3 -1 4 1 1 T: 1 3 3 4 1 9
Sum(S, i)
Sum(S, 7)
int sum(int idx) {
int sum = 0;
while (idx > 0) {
sum += T[idx];
idx -= (idx & -idx); }
return sum;
sum: 0
idx: 7 (0111)

22. Sum(S, i) A: T: Sum(S, i) Sum(S, 7) 1 3 -1 4 1 1 1 3 3 4 1 9 sum: 11 3 -1 4 1 1 T: 1 3 3 4 1 9
Sum(S, i)
Sum(S, 7)
int sum(int idx) {
int sum = 0;
while (idx > 0) {
sum += T[idx];
idx -= (idx & -idx); }
return sum;
sum: 1
idx: 7 (0111)

23. Sum(S, i) A: T: Sum(S, i) Sum(S, 7) 1 3 -1 4 1 1 1 3 3 4 1 9 sum: 51 3 -1 4 1 1 T: 1 3 3 4 1 9
Sum(S, i)
Sum(S, 7)
int sum(int idx) {
int sum = 0;
while (idx > 0) {
sum += T[idx];
idx -= (idx & -idx); }
return sum;
sum: 5
idx: 6 (0110)

24. Sum(S, i) A: T: Sum(S, i) Sum(S, 7) 1 3 -1 4 1 1 1 3 3 4 1 9 sum: 81 3 -1 4 1 1 T: 1 3 3 4 1 9
Sum(S, i)
Sum(S, 7)
int sum(int idx) {
int sum = 0;
while (idx > 0) {
sum += T[idx];
idx -= (idx & -idx); }
return sum;
sum: 8
idx: 4 (0100)

25. Sum(S, i) A: T: Sum(S, i) Sum(S, 7) O(log n) 1 3 -1 4 1 1 1 3 3 4 1 91 3 -1 4 1 1 T: 1 3 3 4 1 9
Sum(S, i)
Sum(S, 7)
int sum(int idx) {
int sum = 0;
while (idx > 0) {
sum += T[idx];
idx -= (idx & -idx); }
return sum;
sum: 8
idx: 0 (0000)
O(log n)

26. Get(S, i) Get(S, i): Sum(S, i) - Sum(S, i-1) Keep a reference to A
int get(int idx) {
int sum = T[idx];
if (idx > 0) {
int z = idx - (idx & -idx);
idx--;
while (idx != z) {
sum -= tree[idx];
idx -= (idx & -idx); }
return sum;
O(log n)

27. Update(S, i,v) 1 3 9 3 4 1 A: 1 3 -1 4 1 4 3 3 7 8 9

28. Update(S, i,v) 1 3 9 3 4 1 A: 1 3 -1 2 4 1 4 3 5 9 10 11

29. Update(S, i,v) 1 3 9 3 2 4 1 A: 1 3 -1 2 4 1 4 3 5 9 10 11

30. Update(S, i,v) 1 3 9 3 2 6 1 A: 1 3 -1 2 4 1 4 3 5 9 10 11

31. Update(S, i,v) A: But how should we encode the index? 1 3 -1 2 4 1 4 31 3 11 3 2 6
But how should we encode the index? 1 A: 1 3 -1 2 4 1 4 3 5 9 10 11

32. Update(S, i, v) And that’s all! Update(S, i, v) O(log n)
void update(int idx ,int val) {
while (idx < |T|) { T[idx] += val;
idx += (idx & -idx); }
O(log n)
And that’s all!

33. Extensions O( 2 𝑑 log 𝑑 𝑛 ) Operations for scaling all values
Operations for finding the index of a particular sum
Multidimensional Binary Indexed Trees
O( 2 𝑑 log 𝑑 𝑛 )
void update(int idx ,int val) {
while (idx <= MAX_VAL) { T[idx] += val;
idx += (idx & -idx); }

34. Extensions O( 2 𝑑 log 𝑑 𝑛 ) Operations for scaling all values
Operations for finding the index of a particular sum
Multidimensional Binary Indexed Trees
O( 2 𝑑 log 𝑑 𝑛 )
void updatey(int x, int y, int val) {
while (y <= MAX_Y) { T[x][y] += val;
y += (y& -y); }
void update(int x, int y, int val) {
while (x <= MAX_X) { updatey(x,y,val);
x += (x & -x); }

35. Problems
EAPC10I – Imagine
BAPC14 – Excellent Engineers …

36. EAPC10I A square board of 1024x1024 square cm
Stickers of 1x1 of party A and B in alternating pattern A B A B B A B A B
Query: how many stickers of each party in (x1,y1,x2,y2)?
Update: a sticker for party A or B was placed at (x,y)

37. Solution Using a 2-dimensional Binary Indexed Tree
How do we store stickers?
We keep track of the sum of stickers for A
Stickers for B = total - #A
How do we query a rectangle (x1,y1,x2,y2)?
Sum(x1,y1,x2,y2) =
Sum(x2, y2) - Sum(x1-1, y2) - Sum(x2, y -1) + Sum(x1-1, y1-1)

38. Questions?