State Representationof

State Space SearchingAlan Tam Siu LungTam@SiuLung.com

9996789196397999

Prerequisite

• State Space Search• Pascal/C/C++

– Familiar with the data types

• Mathematics

Generic GraphSearching Algorithm

• c: container of states• insert start to c• while c not empty

– pop an element from c (with minimum total cost)

– if found goal, return– add/update some elements in c

Generic Graph Searching

• Best first search: choose the node with minimum estimated total path cost = current path cost + estimated cost to goal– Uniform cost search: estimated cost to

goal = 0• Breadth first search: current path cost =

depth of the node in an un-weighted graph

• Depth first search: estimated total path cost = inverse of current path cost

Missionaries and Cannibals

8-puzzle

4 5

6 1 8

7 3 2

1 2 3

4 5 6

7 8

States Storage Problem

• Need to store many states• Memory is limited• Runtime is limited• Compiler is stupid

• we need intelligence to represent states wisely

Operation on States

• 3 operations:– Is goal?– Are these 2 states equal?– Find all neighbors.

• Questions– Can some operation be slow?– Can we pre-compute? (i.e. store

redundant info)– Can states be not uniquely represented?

8-puzzle

• struct state {– int num[9];– int space_pos;

• };

• record state– num : array[1..9] of

0..8– space_pos: 1..9;

• end;

4 5

6 1 8

7 3 2

1 2 3

4 5 6

7 8

Entropy

• Possible Different states: 181440

•Space we used? 320 bits•Using bytes? 80 bits

Missionaries and Cannibals

• struct state {– int m[2], c[2];– bool b;

• };

• record state– m, c : array[1..2] of

1..3;– b : boolean

• end;

Entropy

• Possible Different states: 16•Space we used? 160 bits•Using bytes? 40 bits•One side only? 24 bits

Be realistic

• |{3, 2, 1, 0}|2 |{L, R}| = 32– Store it as M + C * 4 + B * 16

• |{3M3C, 3M2C, 3M1C, 3M, 2M2C, 1M1C, 3C, 2C, 1C, }| |{L, R}| = 20

Computer Organization

• x86 stores whole numbers in:– Binary Form

• x86 stores negative integers in:– 2’s complement

• x86 stores real numbers in:– IEEE 754

• x86 stores alphabets in:– ASCII

Integer Representation

46709394 (10)

= 00000010 11001000 10111010 10010010 (2)

= 0C C8 BA 92 (16) (0x0cc8ba92 in C/C++)

Little Endian:10010010 10111010 11001000 00000010

92 BA C8 02

State Representation

• M + C * 4 + B * 16• M=3, C=3, B=0

– 00001111 (=15)

• M=0, C=0, B=1– 00010000 (=16)

• M=2, C=2, B=1– 00011010 (=26)

If you discovered…

• M=3, C=3, B=0– 00001111 (=15)

• M=0, C=0, B=1– 00010000 (=16)

• M=2, C=2, B=1– 00011010 (=26)

• This is Bitwise Representation

Why bitwise?

• It is the native language of the computer

• It works extremely fast– Integer Multiplication: 4 cycles latency– Moving Bits: 1 cycle latency

• Packing and unpacking is easier to code and thus less error prone

Bitwise Operations

Binary• << shl• >> shr• & and• | or• ^ xor

Unary• ~ not

• shift # bits left• shift # bits right• intersection• union• mutual exclusion

• complement

Examples

1 << 2 4

7 >> 2

13 & 7

12 | 4

12 ^ 24

~55 (8-bit unsigned)

-7 >> 2 (8-bit)

Examples

1 << 2 4

7 >> 2 1

13 & 7 5

12 | 4 12

12 ^ 24 20

~55 (8-bit unsigned) 200

-7 >> 2 (8-bit) -1

Bitmap

• Shift operators and bitwise operators can be used to manage a sequence of bits of 64 elements

• Operations– Get(p : 0..Size-1) : Boolean– Set(p : 0..Size-1)– Unset(p : 0..Size-1)– Toggle(p : 0..Size-1)

Using a 32-bit Integer

• 00000010 11001000 10111010 10010010

• 1 << p = a number with only bit p on• value & (1 << p) != 0 Get(p)• value = value | (1 << p) Set(p)• value = value & ~(1 << p) Unset(p)• value = value ^ (1 << p) Toggle(p)

Bit 31 Bit 0

Using Bitmap

• Space-efficient• Runtime may be faster

– Cache hit– Less copying

• Or maybe slower– More calculations

Massive Calculations

• Example: Count number of bits on a 16-bit integer v

• v = (v & 0x5555) + ((v >> 1) & 0x5555)• v = (v & 0x3333) + ((v >> 2) & 0x3333)• v = (v & 0x0f0f) + ((v >> 4) & 0x0f0f)• v = (v & 0x00ff) + ((v >> 8) & 0x00ff)

Counting #bits

• 1 0 0 1 1 1 1 1 0 0 0 1 0 0 1 0• 01 01 10 10 00 01 00 01• 0010 0100 0001 0001• 00000110 00000010• 0000000000001000

• Exercise: Do it for 32-bit

Massive Calculations

• Given a bit pattern, find a bit pattern which:– only the leftmost bit of a bit

group is on

• E.g.– 1110101101111001 becomes

1000101001000001

Answer

– 1110101101111001 becomes1000101001000001

• The leftmost of a bit group means:– It is 1– Its left is 0

• So:– value = value & ~(value >> 1)

Bit Movements

• Question– 1 1 0 1 0 1 0 0 becomes0101000100010000

• Solution– value = ((value & 0x00f0) << 4)) | (value & 0x000f)– value = ((value & 0x0c0c) << 2)) | (value & 0x0303)– value = ((value & 0x2222) << 1)) | (value & 0x1111)

• Exercise– 1 1 0 1 0 1 0 0 becomes1111001100110000

– 0 1 0 1 becomes0000111100001111

•

Direct Addressing

0000 0000000000000000 1000 1111000000000000

0001 0000000000001111 1001 1111000000001111

0010 0000000011110000 1010 1111000011110000

0011 0000000011111111 1011 1111000011111111

0100 0000111100000000 1100 1111111100000000

0101 0000111100001111 1101 1111111100001111

0110 0000111111110000 1110 1111111111110000

0111 0000111111111111 1111 1111111111111111

Complex Bit Patterns

• Othello– 11101011 (where occupied)– 10001001 (color)– You are 1, where you can play?

• Solution– 11101011 & ~10001001 = 01100010– 11101011 + 01100010 = 01001101– 01001101 & ~11101011 = 00000100

• How about the reverse direction?

Other Uses

• Store multiple values, each one assigned a bit-range

• (value % 32) == (value & 31)• Storing sets of 64 elements

– Union– Difference– Is Subset

Multiple Comparisons

switch (i) {case 1:case 3:case 4:case 7:case 8:case 10:

f();break;

default:g();

}

if ((1<<i) & 0x039a)f();

elseg();

Lexicographical Ordering

• Consider the 8 puzzle• Store it verbatim: 89 = 387420489

• Hash table? How large?• Calculate a mapping from the

362880 possible permutations to 0..362879

Example

• What are smaller than 530841672?– 5308416[0-6]?: 1 1– 530841[0-5]??: 1 2– 53084[0-0]???: 0 6– 5308[0-3]????: 2 24– 530[0-7]?????: 5 120– 53[^]??????: 0 720– 5[0-2]???????: 3 5040– [0-4]????????: 5 40320