https://wiki.haskell.org/index.php?title=Talk:Battleship_game_combinatorics&feed=atom&action=historyTalk:Battleship game combinatorics - Revision history2017-03-01T17:40:05ZRevision history for this page on the wikiMediaWiki 1.19.14+dfsg-1https://wiki.haskell.org/index.php?title=Talk:Battleship_game_combinatorics&diff=54631&oldid=prevJestingrabbit at 19:59, 9 November 20122012-11-09T19:59:49Z<p></p>
<table class='diff diff-contentalign-left'>
<col class='diff-marker' />
<col class='diff-content' />
<col class='diff-marker' />
<col class='diff-content' />
<tr valign='top'>
<td colspan='2' style="background-color: white; color:black;">← Older revision</td>
<td colspan='2' style="background-color: white; color:black;">Revision as of 19:59, 9 November 2012</td>
</tr><tr><td colspan="2" class="diff-lineno">Line 2:</td>
<td colspan="2" class="diff-lineno">Line 2:</td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>For the actual game configuration, with ships of length 5, 4, 3, 3 and 2, with length 3 ships distinct, I get 30,093,975,536. As a lower bound for that bound for that problem, consider that a ship of length m can eliminate at most (n+1)(m+1)-2 placements of the length n ship. Placing the ships in order 5, 4, 3, 3, 2, there are 120 placements for the 5, at least 140-28=112 for the 4, at least 160-22-18=120 for the first 3, at least 160-22-18-14=106 for the second 3 and 180-16-13-10-10=131 for the 2. This gives a lower bound of 120*112*120*106*131=22395340800. That's for distinct 3s, but this gives a lower bound of more than 5 times your answer. [[User:Jestingrabbit|Jestingrabbit]] 08:39, 9 November 2012 (UTC)</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>For the actual game configuration, with ships of length 5, 4, 3, 3 and 2, with length 3 ships distinct, I get 30,093,975,536. As a lower bound for that bound for that problem, consider that a ship of length m can eliminate at most (n+1)(m+1)-2 placements of the length n ship. Placing the ships in order 5, 4, 3, 3, 2, there are 120 placements for the 5, at least 140-28=112 for the 4, at least 160-22-18=120 for the first 3, at least 160-22-18-14=106 for the second 3 and 180-16-13-10-10=131 for the 2. This gives a lower bound of 120*112*120*106*131=22395340800. That's for distinct 3s, but this gives a lower bound of more than 5 times your answer. [[User:Jestingrabbit|Jestingrabbit]] 08:39, 9 November 2012 (UTC)</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div><ins style="color: red; font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div><ins style="color: red; font-weight: bold; text-decoration: none;">Ah, I see what's going on. I'll edit the page to reflect what's happening. [[User:Jestingrabbit|Jestingrabbit]] 19:59, 9 November 2012 (UTC)</ins></div></td></tr>
</table>Jestingrabbithttps://wiki.haskell.org/index.php?title=Talk:Battleship_game_combinatorics&diff=54625&oldid=prevJestingrabbit: New page: It seems like this table is really wrong. If you look at the second entry, the number of ways to place 2 ships of length 2, there are 8 ways to put the first in a corner which blocks 4 mov...2012-11-09T08:39:07Z<p>New page: It seems like this table is really wrong. If you look at the second entry, the number of ways to place 2 ships of length 2, there are 8 ways to put the first in a corner which blocks 4 mov...</p>
<p><b>New page</b></p><div>It seems like this table is really wrong. If you look at the second entry, the number of ways to place 2 ships of length 2, there are 8 ways to put the first in a corner which blocks 4 moves for the second ship, 28 ways to place it along a side which blocks 5, 32 ways to place it perpendicular to a side which blocks 6 moves, and 112 ways to place the first ship in the middle of the board, which block 7 moves for the second ship ie there are (8*176 + 28 * 175 + 32 * 174 + 112 * 173)/2 = 15626 ways to place 2 indistinguishable length 2 ships, and the table says there are 13952.<br />
<br />
For the actual game configuration, with ships of length 5, 4, 3, 3 and 2, with length 3 ships distinct, I get 30,093,975,536. As a lower bound for that bound for that problem, consider that a ship of length m can eliminate at most (n+1)(m+1)-2 placements of the length n ship. Placing the ships in order 5, 4, 3, 3, 2, there are 120 placements for the 5, at least 140-28=112 for the 4, at least 160-22-18=120 for the first 3, at least 160-22-18-14=106 for the second 3 and 180-16-13-10-10=131 for the 2. This gives a lower bound of 120*112*120*106*131=22395340800. That's for distinct 3s, but this gives a lower bound of more than 5 times your answer. [[User:Jestingrabbit|Jestingrabbit]] 08:39, 9 November 2012 (UTC)</div>Jestingrabbit