{"id":555,"date":"2014-12-30T07:32:05","date_gmt":"2014-12-30T12:32:05","guid":{"rendered":"http:\/\/dlewis.net\/blog\/?page_id=555"},"modified":"2014-12-30T07:32:05","modified_gmt":"2014-12-30T12:32:05","slug":"the-twelve-coins","status":"publish","type":"page","link":"https:\/\/dlewis.net\/blog\/the-twelve-coins\/","title":{"rendered":"The Twelve Coins"},"content":{"rendered":"

1) The Heavier Coin<\/strong><\/p>\n

You have a balance and twelve coins, one of which is heavier than the rest, but you don’t know which one it is. Using just 3 weighings on the balance, can you identify which coin is the odd one out?<\/p>\n

SOLUTION<\/span><\/p>\n

This one is pretty easy. There\u00a0are a few ways to solve this but they all follow the same principle and pattern.\u00a0I\u2019m going to pick one approach — probably the<\/span> most straight-forward one.<\/p>\n

Step 1: Put four coins on either side of the<\/span> balance, reserving four others. If the\u00a0<\/span>balance stays even, you know that the<\/span> heavier<\/span> coin<\/span> is among the<\/span> four reserved. If the b<\/span>alance tips, you know that the<\/span> heavier<\/span> coin<\/span> is on the<\/span> side that went down. Either way, you now have isolated the<\/span> heavier<\/span> coin<\/span> within a subset of four.<\/p>\n

Step 2: Put two coins of the<\/span> four identified in Step 1 on either side of the<\/span> balance, reserving none. Whichever side goes down has the<\/span> heavier<\/span> coin<\/span>.<\/p>\n

Step 3: You have two coins left and can use the<\/span> scale once more. That\u2019s a great situation to be in! Put one coin<\/span> on either side and whichever sides drops contains the<\/span>heavier<\/span> coin<\/span>. You win.<\/p>\n

Pretty simple, right? The<\/span> trick is to realize that if the<\/span> balance stays level, you\u2019ve still gained key information.<\/p>\n

2) The Different Coin<\/strong><\/p>\n

You have a balance and twelve coins, one of which weighs a different amount than the rest, but you don’t know which one it is — and you don\u2019t know if the different coin is heavier or lighter than the rest. Using just 3 weighings on the balance, can you identify which coin is the odd one out — and whether it’s heavier and lighter than the others?<\/p>\n

SOLUTION<\/span><\/p>\n

Here\u2019s\u00a0a framework for thinking through the<\/span> solution. (The<\/span> full solution is pages and pages long.)<\/p>\n

First, let\u2019s arbitrarily label all the coins. 1 through 9 works for most, and let\u2019s use T, J, and Q for ten, eleven, and twelve. (That\u2019s Ten, Jack, Queen, if you want to know why I chose those letters.)<\/p>\n

Second, let\u2019s put coins 1-2-3-4 on one side of the balance and coins 5-6-7-8 on the other, keeping 9-T-J-Q in reserve. One of three things will happen: The 1-2-3-4 side will go down; the 1-2-3-4 side will go up; or the balance will stay level. For sake of example, let\u2019s say that it stays level. Good news: you know that the odd coin out is one of 9-T-J-Q. You\u2019ve eliminated two-thirds of the possible answers. Had the balance tipped, you\u2019d have only eliminated a third.<\/p>\n

The bad news is that if you follow your instinct and balance 9-T versus J-Q, you\u2019re not able to eliminate any more coins — it\u2019s going to tip. Balancing, say, 9 versus T has a similar problem — if it doesn\u2019t tip, you\u2019re left with J and Q, and putting both on the balance tells you very little.<\/p>\n

There are three things you have to realize in order to get this one done.<\/p>\n

1) If a coin goes up in one weighing and down in another, it\u2019s not the coin you\u2019re looking for.<\/p>\n

2) You\u2019re going to\u00a0need to test an odd numbers of coins, so you\u2019ll need to add in a coin of known weight (that is, one that\u2019s already been identified as one of the 11 normals) at some point. (As it turns out, you\u2019ll always<\/em> need to test an odd number of coins at one point or another, making this critical.)<\/p>\n

3) The first weighing isn\u2019t about eliminating coins. It\u2019s about collecting information. The second is similar, but more importantly, about getting down to no more than three choices left, and knowing something about each of those choices.<\/p>\n

Let\u2019s say that the 6 coin is the odd coin out and you\u2019ll see what I mean. Again, et\u2019s put coins 1-2-3-4 on one side of the balance and coins 5-6-7-8 on the other, keeping 9-T-J-Q in reserve for our first weighing. The 1-2-3-4 side goes up and the 5-6-7-8 side goes down. So we know that, if the odd coin is one of the 1-2-3-4, it\u2019s lighter than the rest; and if the odd coin is in the 5-6-7-8 group, it\u2019s heavier than the rest. The 9-T-J-Q? All normals.<\/p>\n

Second weighing: Put 1-2-5 on the left. On the right, put 3-6-Q. Note that the 4, 7, and 8 aren\u2019t being tested again and the Q is being used as a placeholder, to make sure the scale can balance in case the 4, 7, or 8 are the odd duck. The right side goes down again, meaning that the if the coin is the 1, 2, or 5, it\u2019s lighter than the rest; or if it\u2019s the 3, 6, or Q, it\u2019s heavier. We can remove the Q from the solution set because it\u2019s already been proven to be a normal coin from the first weighing. We can also remove the 3 and 5 because they can\u2019t be heavy in one weighing and light in the other. They\u2019re both normals.<\/p>\n

So we\u2019re left with 1, 2, and 6. We know if the answer is 1 or 2, that\u2019s because it\u2019s too light. We know if the answer is 6, it\u2019s because it\u2019s too heavy. Our third weighing: 1 versus 2. If it tips, whichever side goes up holds the mystery coin. If it stays balanced, the mystery coin is the too-heavy 6. That\u2019s what happens and we\u2019ve identified our coin.<\/p>\n

Let\u2019s do this again, starting with step 2, but this time, let\u2019s make the 4 coin lighter than the other eleven. To recap, the 4 was on the left side of the balance for the first weighing but not on the balance at all for the second. So this time, the balance stays level on the second weighing. We now know that our mystery coin is one of the 4, 7, and 8. If the answer is 7 or 8, it\u2019s because the coin is too heavy, so we weight 7 versus 8. Had it tipped, whichever side goes down contains the maladjusted coin. Otherwise, the balance will stay level and the 4 is our answer.<\/p>\n

You can try other permutations if you\u2019d like. The biggest trick is to make sure you use a dummy coin in the second weighing to make sure that the third weighing has only three options, max, to test.<\/p>\n","protected":false},"excerpt":{"rendered":"

1) The Heavier Coin You have a balance and twelve coins, one of which is heavier than the rest, but you don’t know which one it is. Using just 3 weighings on the balance, can you identify which coin is the odd one out? SOLUTION This one is pretty easy. There\u00a0are a few ways to […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":[],"_links":{"self":[{"href":"https:\/\/dlewis.net\/blog\/wp-json\/wp\/v2\/pages\/555"}],"collection":[{"href":"https:\/\/dlewis.net\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/dlewis.net\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/dlewis.net\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/dlewis.net\/blog\/wp-json\/wp\/v2\/comments?post=555"}],"version-history":[{"count":1,"href":"https:\/\/dlewis.net\/blog\/wp-json\/wp\/v2\/pages\/555\/revisions"}],"predecessor-version":[{"id":556,"href":"https:\/\/dlewis.net\/blog\/wp-json\/wp\/v2\/pages\/555\/revisions\/556"}],"wp:attachment":[{"href":"https:\/\/dlewis.net\/blog\/wp-json\/wp\/v2\/media?parent=555"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}