a*****g
发帖数: 19398 | 1 #!/usr/bin/env pike
// legal.pike - Count the number of legal go boards.
// Copyright 2005 by Gunnar Farneb?ck
// [email protected]
/* */
//
// You are free to do whatever you want with this code.
//
//
// This program computes the number of legal go board configurations
// for a rectangular board of a given size. It is efficient enough to
// handle boards up to 8x8 within minutes and up to 11x11 in less than
// 24 hours (on a fast computer). For rectangular boards MxN, large N
// can be handled efficiently as long as M is kept small.
//
// The program is inspired by the Computer Go List postings of Jeffrey
// Rainy,
//
// Is there any published articles or websites stating the exact number of
// legal goban configurations that are possible for a given goban size ?
//
// The best I found, from Sensei's Library is :
//
// 1x1: 1 legal, 2 illegal, prob 0.333333
// 2x2: 57 legal, 24 illegal, prob 0.703704
// 3x3: 12675 legal, 7008 illegal, prob 0.643957
// 4x4: 24318165 legal, 18728556 illegal, prob 0.564925
// 4x5: 1840058693 legal, 1646725708 illegal, prob 0.527724
//
// I realise this is of no interest except from a purely
// theoretical/combinatorial viewpoint. However, I think I could come up
with
// the exact values up to at least 9x9. Was this done before ? Would
anyone
// find it useful or at least somehow interesting ?
//
// and the response from Eric Boesch,
//
// >Would anyone find it useful or at least somehow interesting ?
//
// Yes, I think the answer would be interesting, and the method used to
// arrive at numbers for such large boards would be even more
// interesting. At first I thought you were crazy to think the
// computation was practical for 9x9, but after considerable thought, I'm
// not so sure. I think I've finally worked out an angle. In any case, it
// sounds like not only did you find an angle first, but you also had the
// insight to realize an answer might be possible in the first place,
// which I did not. Considering where you say Sensei's left off (5x4), I
// bet they used brute force, which isn't very interesting.
//
// Here are my thoughts on the problem. I have not actually worked out
// the solution -- I figure you have dibs on this problem anyway :)
// Possible spoiler?
//
// Bisection and divide-and-conquer, while (I think) straightforward
// enough for 1-dimensional boards, seemed very difficult to do for
// 2-dimensional ones. Hmm...
//
// Instead of divide-and-conquer, I think you should work out from the
// corner, adding stones one at a time, column by column, keeping an
// array mapping all possible border states to the number of different
// overall region states that share that same border state. The border is
// the set of intersections that are inside the region that are adjacent
// to intersections outside the region. The border state is not only the
// color of the border intersections, but also whether those stones
// already have liberties, or whether they need to connect to outside
// liberties, and also whether those stones are connected to each other.
// For example:
//
// -----
// | #
// | # O
// | # #
// | # O
// | O .
// -----
//
// If this is our region, then the border consists of the four rightmost
// intersections (the rightmost ones in each row). The border colors are
// #,O,#,O, and "." (empty), but the border state is more than just the
// colors of the border stones. It's also important to know that the
// topmost O stone needs to connect (directly or indirectly) to an
// outside liberty. The two # stones on the border also need to connect
// to a liberty, but since they are already connected to each other, a
// single liberty at either end suffices.
//
// So the number of possible border configurations is strictly greater
// than 3^(# of border intersections). I have no idea what the exact
// number would be, but as long as the effective exponent base isn't too
// big, the space demands for 9x9 should be manageable. For example, 4^9
// is just a quarter million.
//
// In sum, when you're adding a new intersection to the region, you
// iterate through all of the old border states, and through all of the
// possible colors for that new intersection (black, white, empty). For
// each old border state, you compute the new border state that results
// from adding the given stone to the old border, and you add the number
// of positions having the old border state to the number of positions
// having the new border state. Once you have added all 81 stones, the
// single (null) border state number will equal the total number of valid
// positions -- make sure to use either floating point or bignums.
//
// For the full postings, see
// http://computer-go.org/pipermail/computer-go/2005-January/002387.html
// http://computer-go.org/pipermail/computer-go/2005-January/002412.html
//
//
// This program implements the algorithm as outlined by Eric.
// It is written in Pike. See http://pike.ida.liu.se for documentation
// and download. It should be noted that the "int" type in Pike
// automatically switches to bignums when the values become too large
// for the native integers on the platform.
//
// The internal representation of the border state is a string with a
// length equaling the height of the board and containing the
// following characters:
//
// X: A black stone with at least one liberty guaranteed.
// O: A white stone with at least one liberty guaranteed.
// .: An empty vertex.
// |: An edge vertex, only used initially while traversing the first column.
// a,b,c, etc.: White strings with no liberty found so far.
// A,B,C, etc.: Black strings with no liberty found so far.
//
// For strings without liberties the rule is that connected stones
// have the same letter and disconnected stones have different
// letters. Furthermore the state is normalized so that the first
// occurence of "a" always precedes the first occurence of "b", etc.,
// and so that there are no holes in the letter sequences. This
// normalization is not required for the program to give correct
// results (*) but improves the performance by keeping the set of
// border states as small as possible.
//
// There is clearly no need to keep track of connectivity for stones
// with liberties.
//
// The state is recorded from the bottom and up. The position in
// Eric's example above is represented by the state string ".OAaA".
//
// During the computation of new state the state string is converted
// into an array of single-character strings. This is because strings
// are immutable in Pike but arrays can be destructively modified.
//
// (*) Not quite true since one piece of the code would need to be
// modified in this case, as documented in the code comments.
//
//
// Results for quadratic boards computed by this program:
//
// 1x1 1
// 2x2 57
// 3x3 12675
// 4x4 24318165
// 5x5 414295148741
// 6x6 62567386502084877
// 7x7 83677847847984287628595
// 8x8 990966953618170260281935463385
// 9x9 103919148791293834318983090438798793469
// 10x10 96498428501909654589630887978835098088148177857
//
// The time complexity seems to increase roughly by 7^N which means
// that 11x11 should be solvable in about 24 hours, 12x12 in a week,
// and 13x13 in less than two months. Please report the results if you
// find the computer time to do this.
/**********************************************************/
// Normalize the state array so that the first occurences of strings
// in "state" that are included in "strings" are ordered and without
// holes. If called succesively with black_strings and white_strings
// the state will be normalized as described above.
array(string) normalize_state(array(string) state, array(string) strings)
{
int j = 0;
for (int i = 0; i < sizeof(state); i++) {
if (has_value(strings, state[i])) {
if (state[i] > strings[j]) {
// Out of order or a hole in the sequence. Exchange
// this letter with the next letter in turn. Use "@"
// as a temporary placeholder.
string x = state[i];
state = replace(state, strings[j], "@");
state = replace(state, x, strings[j]);
state = replace(state, "@", x);
j++;
}
else if (state[i] == strings[j])
j++;
}
}
return state;
}
// Count the number of legal boards of the given size.
int count_legal_boards(int height, int width)
{
// The border state count is represented by a mapping which
// associates each border state string with the number of
// configurations of the stones placed on the board so far having
// that border state.
mapping(string:int) new_state_count = ([]);
mapping(string:int) old_state_count = ([]);
// Letters used for strings without liberties. There can be at
// most height/2 strings of each color in the border state but we
// need one extra letter to avoid indexing out of range in
// normalize_state().
//
// The division by "" (the empty string) splits the string into an
// array of one-character strings.
//
// Naturally "O" and "X" are excluded here. If someone in the far
// future finds a computer capable of handling heights larger than
// 44 fast enough to be useful, these lists need to be extended. Any
// characters can be used in arbitrary order as long as there are no
// collisions.
array(string) white_strings = "abcdefghijklmnpqrstuvwyz"[0..height/2+1] /
"";
array(string) black_strings = "ABCDEFGHIJKLMNPQRSTUVWYZ"[0..height/2+1] /
"";
// The initial state is "|" repeated height times, e.g. "|||||"
// for 5xN boards.
string edge = "|" * height;
new_state_count[edge] = 1;
// Keep track of the maximum number of border states for statistics.
int max_number_of_border_states = 0;
// Traverse the board in the order new vertices are added. The outer
// loop goes from the left to the right and the inner loop from the
// bottom to the top.
for (int j = 0; j < width; j++)
for (int i = 0; i < height; i++) {
// Move the previous new state count to old_state_count and
// start over with an empty new_state_count.
old_state_count = new_state_count;
new_state_count = ([]);
// Loop over the previous states.
foreach (indices(old_state_count), string state)
// Add an empty vertex, a black stone, and a white stone in turn.
foreach (({"empty", "black", "white"}), string new_stone) {
// Convert the state string to a state array.
array(string) new_state = state / "";
// What matters in the computation of the new state is what
// we have to the left and below. If we are at the bottom of
// the column the neighbor below is the edge symbol "|".
string left = new_state[i];
string down = (i == 0) ? "|" : new_state[i-1];
// Of particular interest is whether we have a string
// without liberties of one of the colors to the left or
// below.
int black_string_left = has_value(black_strings, left);
int black_string_down = has_value(black_strings, down);
int white_string_left = has_value(white_strings, left);
int white_string_down = has_value(white_strings, down);
// If we find that the new configuration is illegal we set
// this variable to 1 to mark that it should be discarded.
int bad_state = 0;
// Handle each addition case by case.
switch (new_stone) {
case "empty":
// When we add an empty vertex, the configuration is
// guaranteed to be valid and the new state at the current
// position will definitely be ".". If we have a string
// without liberties to the left or below it has now
// received a liberty and its stones should be converted
// either to "X" or "O".
new_state[i] = ".";
if (black_string_left)
new_state = replace(new_state, left, "X");
if (black_string_down)
new_state = replace(new_state, down, "X");
if (white_string_left)
new_state = replace(new_state, left, "O");
if (white_string_down)
new_state = replace(new_state, down, "O");
break;
case "black":
// If we have a white string without liberties to the left
// and this was its last stone on the border, it can no
// longer get any liberty and thus the configuration is
// illegal.
if (white_string_left && sizeof(state - left) == height - 1) {
bad_state = 1;
break;
}
// If we have at least one empty vertex or a black stone
// with liberty to the left or to the right, the new stone
// will also have at least one liberty.
if (has_value("X.", left) || has_value("X.", down)) {
new_state[i] = "X";
// Furthermore, if the other neighbor was a black string
// without liberties, it has also received a liberty
// now.
if (black_string_left)
new_state = replace(new_state, left, "X");
else if (black_string_down)
new_state = replace(new_state, down, "X");
}
else if (black_string_left && black_string_down) {
// Both neighbors are black strings without liberties.
// These need to be merged and the new string will also
// lack liberties. There is no need to set the state at
// the current position explicitly as it will be part of
// the string inherited from the left.
if (left != down)
new_state = replace(new_state, down, left);
}
else if (black_string_down) {
// Black string without liberties below and a white
// stone (with or without liberties) or the edge to the
// left. Extend the string below to the current
// position.
new_state[i] = down;
}
else if (!black_string_left) {
// If we have a black string without liberties to the
// left and a white stone or the edge below we do not
// need to do anything as the border state remains
// unchanged by adding the new stone to the string to
// the left.
//
// If we have white stones or edges both to the left and
// below we get a new string without liberties now. We
// temporarily use the last available string letter,
// which is guaranteed to be free and let the state
// normalization change it to the appropriate letter.
// (However, if state normalization were to be turned
// off it would be necessary to search for a free string
// letter here.)
new_state[i] = black_strings[-1];
}
break;
case "white":
// This code is identical to the "black" case above but
// with reversed roles for black and white. We do not
// repeat the comments above.
if (black_string_left && sizeof(state - left) == height - 1) {
bad_state = 1;
break;
}
if (has_value("O.", left) || has_value("O.", down)) {
new_state[i] = "O";
if (white_string_left)
new_state = replace(new_state, left, "O");
else if (white_string_down)
new_state = replace(new_state, down, "O");
}
else if (white_string_left && white_string_down) {
if (left != down)
new_state = replace(new_state, down, left);
}
else if (white_string_down)
new_state[i] = down;
else if (!white_string_left)
new_state[i] = white_strings[-1];
break;
}
// Throw away bad configurations. Normalize good ones and
// then add them to the state count.
if (!bad_state) {
new_state = normalize_state(new_state, black_strings);
new_state = normalize_state(new_state, white_strings);
new_state_count[new_state * ""] += old_state_count[state];
}
}
// Update statistics.
if (sizeof(new_state_count) > max_number_of_border_states)
max_number_of_border_states = sizeof(new_state_count);
}
// The board has been traversed. The final border states which
// include black or white strings without liberties correspond to
// illegal board configurations and must be excluded. We do this by
// summing the state counts for state strings only containing the
// characters ".", "X", and "O".
int sum = 0;
foreach (indices(new_state_count), string state) {
if (sizeof(state - "." - "X" - "O") == 0)
sum += new_state_count[state];
}
// Print statistics.
write("Max number of border states: %d\n",
max_number_of_border_states);
return sum;
}
int main(int argc, array(string) argv)
{
if (argc < 3) {
werror("Usage: pike legal.pike height width\n");
exit(1);
}
int height = (int) argv[1];
int width = (int) argv[2];
int num_legal = count_legal_boards(height, width);
// If the board is too large, we cannot convert to float before the
// division to compute the fraction of legal boards since that would
// cause overflow. With this trick we use bignum integers in the
// division and get a result that is safe to convert to float.
write("%dx%d: %d (%2.4f%%) legal boards\n", height, width, num_legal,
0.000001 * (100000000 * num_legal / pow(3, height*width)));
// Signal successful execution.
return 0;
} | h**********c
发帖数: 4120 | 2 我搜索了键单词sym,没有看到属于对称性的文档话 | a*****g
发帖数: 19398 | 3 不是特别需要排除对称
【在 h**********c 的大作中提到】
: 我搜索了键单词sym,没有看到属于对称性的文档话
| h**********c
发帖数: 4120 | 4 哎呀，你这个题，不好办啊！
我老提供点inuit.
你说比如算了n=3，下一层归纳加Markov铁链变换，是不是可以分布？
能拿到蒜肠奖
而且算偶数有点无厘头
【在 a*****g 的大作中提到】
: 不是特别需要排除对称
| r****y
发帖数: 26819 | 5 一看最初的几个例子就知道这个程序没有排除对称的局面。
【在 h**********c 的大作中提到】
: 我搜索了键单词sym,没有看到属于对称性的文档话
| h**********c
发帖数: 4120 | 6 re
【在 r****y 的大作中提到】
: 一看最初的几个例子就知道这个程序没有排除对称的局面。
| r****y
发帖数: 26819 | 7 比如，2X2的棋盘，考虑对称，合法的局面绝没有57个。
全空：1
1黑：1
2黑：2
3黑：1或者2，看考虑哪种对称
1白：1
2白：2
3白：1或2
1白1黑：1或者2
1白2黑：1或者2
2白1黑：1或者2
【在 h**********c 的大作中提到】
: re
| h**********c
发帖数: 4120 | 8 钢材跑步的时候想归纳是不能全从合法的状态开始的，
再议 | a*****e
发帖数: 1700 | 9 pike... 好熟悉的名字，十五年前经常用，没想到现在还在呢 | h**********c
发帖数: 4120 | 10 想着想着往非博那基的方向走，太复杂了
赶紧醒了
【在 h**********c 的大作中提到】
: 钢材跑步的时候想归纳是不能全从合法的状态开始的，
: 再议
| h**********c
发帖数: 4120 | 11 设i=m，我们计算了全部合法状态Q，并且计算了一些不合法的状态NQ_1,NQ_1只有在相
邻位置放１子且需要在m+1层放子,这有点想grammar里面的lookahead.
可以把Q和 NQ_1全部状态分布，计算i+1.
所以棋盘的道数应该是
1
3
5
...
不应该是偶数，暂不考虑对称性
可能还是不complete
想着想着往非博那基的方向走，太复杂了
赶紧醒了
【在 h**********c 的大作中提到】
: 想着想着往非博那基的方向走，太复杂了
: 赶紧醒了
|
|
