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--- chess:programming:endgame_table-bases:encoding [2022/01/11 16:07] – peter
+++ chess:programming:endgame_table-bases:encoding [2022/01/11 16:18] (current) – peter
@@ Line 59: / Line 59: @@
   * For example KRRvK.
     * We do not want to have one index value for R1 on a1, R2 on b1 and another index value for R1 on b1, R2 on a1.
-    * What we do is place the two Rs **together**.
+    * We can place two Rs "together" in 62*61/2 ways.
-    * If we have 62 squares left, we can place two Rs "together" in 62*63/2 ways.
+    * If we have 3 Rs, it is 62*61*60/6 ways.
-    * If we have 3 Rs, it is 62*63*64/6 ways.
+    * If we have 4 Rs, it is 62*61*60*59/24 ways."
-    * If we have 4 Rs, it is 62*63*64*65/24 ways.
 </WRAP>
@@ Line 116: / Line 115: @@
 There are special case routines for indexing KKKvNN for suicide chess.
+----
+===== Line by line explanation of the "111" case =====
+We are calculating an index for wK, wR, bK (pos[0], pos[1], pos[2]).
+  * We do not make use of the fact that the two Ks cannot be on adjacent squares.
+    * (So this also works for indexing wQ, wR, wB, for example.)
+The wK has been mapped to the a1-d1-d4 triangle.
+  * If wK is on the diagonal, then wR is on or below the diagonal.
+  * If wK and wR are on the diagonal, then bK is on or below the diagonal.
+<code>
+i = pos[1] > pos[0];
+int j = (pos[2] > pos[0]) + (pos[2] > pos[1]);
+</code>
+Precalculate the values we will have to deduct from pos[1] and pos[2].
+<code>
+if (OffdiagA1H8[pos[0]])
+  idx = Triangle[pos[0]] * 63*62 + (pos[1] - i) * 62 + (pos[2] - j);
+</code>
+wK is below the diagonal.
+  * Triangle[pos[0]] maps the b1-d1-d3 triangle to 0...5.
+  * There are 63 squares for wR (pos[1] - i is in 0...62) and 62 squares for bK (pos[2] - j is in 0...61).
+  * Here, idx wil be in 0...(6*63*62-1).
+<code>
+else
+if (OffdiagA1H8[pos[1]])
+  idx = 6*63*62 + Diag[pos[0]] * 28*62 + Lower[pos[1]] * 62 + pos[2] - j;
+</code>
+wK is on the half-diagonal a1-d4, wR is below the diagonal.
+  * Diag[pos[0]] maps a1-d4 to 0...3.
+  * Lower[pos[1]] maps the b1-h1-h7 triangle to 0..27. 62 squares remain for bK.
+  * Here, idx will be in (6*63*62)...(6*63*62 + 4*28*62-1).
+<code>
+else
+if (OffdiagA1H8[pos[2]])
+  idx = 6*63*62 + 4*28*62 + (Diag[pos[0]]) * 7*28 + (Diag[pos[1]] - i) * 28 + Lower[pos[2]];
+</code>
+Now wK is on the half-diagonal a1-d4, wR is on the diagonal a1-h8, bK is below the diagonal.
+<code>
+else
+  idx = 6*63*62 + 4*28*62 + 4*7*28 + (Diag[pos[0]] * 7*6) + (Diag[pos[1]] - i) * 6 + (Diag[pos[2]] - j);
+</code>
+And the final case: all on the diagonal a1-h8.
+----
+==== References ====
+http://www.talkchess.com/forum3/viewtopic.php?t=59947
+http://www.talkchess.com/forum3/viewtopic.php?t=59904