## Chocolate Bunnies

A candy maker manufactures 7 Chocolate Easter bunnies of increasing sizes. His plan was to use 10 grams of chocolate for the first bunny he made and 10 additional grams for each of the successive ones. So, at the end, the bunnies should weigh 10, 20, 30, 40, 50, 60, and 70 grams respectively. However, after making the first k bunnies, he takes a lunch break and during that time a mischievous helper messes up his scale so at the end, the remaining 7-k bunnies wind up weighing 10 grams more than they should. Suppose that you are to determine when he went out to lunch by means of a [balance] scale with 3 outputs; less than, equal, and greater than.
• What is the minimum number of weighings you need to determine k?
• Construct a scheme of weighings which achieves this minimum.
• What if there are originally eight bunnies (10 through 80)?
Here are a few clarifications from reader's questions: There are only seven bunnies, made in order (10, 20, etc.). 'k' can be any number from 0 to 7.

Source: Univ of Calif, San Diego, Math 187 (Cryptography) Puzzle #3

Solutions were received from Denis Borris, Larry Baum, Craig Gentry, Yaacov Weiss.

Craig Gentry solved both problems succinctly:

```7 bunnies, 2 weighings
1) Weigh 3 and 4 on the left against 7 on the right (So we're on the
same page, bunny 3 is the one that weighs 30 or 40.)
2) If left is heavier (and 1, 2, or 3 is the first heavy bunny),
then weigh 1 and 2 against 3.  If left is heavier, 1 is the first heavy
bunny.  If balanced, 2.  If right, 3.
2) If right is heavier (and 5, 6, or 7 is the first heavy bunny),
then weigh 5 and 6 against 7 and 4.  If left heavier, 5 is the first.
If balanced, 6.  If right, 7.
2) If balanced (and 4 or 8 is the first heavy bunny (Saying 8 is the
first heavy bunny is just shorthand for saying that none of the 7
bunnies is heavy)), weigh 4 against 1 and 3.  If left is heavier, then 4
is the first heavy bunny; otherwise, there is no heavy bunny.

8 bunnies, 2 weighings
Exactly the same as above, except if the first weighing (3 and 4
against 7) is balanced, we have 3 possibilities remaining (4, 8 and 9),
and we weigh 4, 5 and 6 against 7 and 8.  If left heavier, 4 is the
first heavy bunny.  If balanced, no heavy bunny.  If right, 8 is the
first heavy bunny.

In general, you can do up to 3^n - 1 bunnies in n weighings.  I'll
explain how if there's any interest.  This does not apply to the
trivial cases of 1, 2, or 3 bunnies, which cannot be completely
solved in any number of weighings.
-Craig Gentry   ts208@is5.nyu.edu
```
Other Solutions:
1. For the first problem of 7 bunnies, 2 weighings are needed. Denis Borris sent:
```1- weigh B6 against B1 B2 B3
3 cases will be even
3 cases will be B6 heavy
2 cases will be B6 light

2- 3 cases even;
B3 B4 against B7

3- 3 cases B6 heavy;
B4 B5 against B2 B7

4- 2 cases B6 light;
B1 B2 against B3
```
Larry Baum assumed more than one of each bunny was available. Although that's not what I originally intended, his solution has the advantage of being able to determine both weighings prior to starting:
```Notation:
117 means two of bunny size 1 and one of bunny size 7

L1 = weight of left side on weighing 1
R1 = weight of right side on weighing 1
L2 = weight of left side on weighing 2
R2 = weight of right side on weighing 2

<= means L1 means L1R2
etc.

Weighing #1: 117 vs 45

Weighing #2: 1235 vs 56

k    L1   R1   L2   R2  result
0   120  110  150  130  >>
1   100  110  140  130  <>
2   100  110  130  130  <=
3   100  110  120  130  <<
4   100  100  120  130  =<
5   100   90  110  120  ><
6   100   90  110  110  >=
7    90   90  110  110  ==
```
Yaacov Weiss sent:
```(My unit is a 10gram unit - now i work with single digits.)

First put bunnies 167 on one side and 2345 on the other.
given k, all bun's up to k are the correct size,
while the ones BIGGER than k are too big.

K       1,6,7     <=>     2,3,4,5
0       2+7+8=17  <    18=3+4+5+6
1       1+7+8=16  <    18=3+4+5+6
2       1+7+8=16  <    17=2+4+5+6
3       1+7+8=16  =    16=2+3+5+6
4       1+7+8=16  >    15=2+3+4+6
5       1+7+8=16  >    14=2+3+4+5
6       1+6+8=15  >    14=2+3+4+5
7       1+6+7=14  =    14=2+3+4+5

if Result is '='
6 ------- 123
K=3   7     >    6
K=7   6     =    6

else if Result is '<'
3 ------- 12
K=0   4    <    5
K=1   4    =    4
K=2   4    >    3

else (Result is '>')
47 ------- 56
K=4  12    <    13
K=5  12    =    12
K=6  12    >    11

As we see - whatever result we get after the first weighing
we can solve it on the second weighing.
```
2. For 8 bunnies, still only 2 weighings are needed. Denis Borris sent:
```VARIABLE NAMES:
Bunnies in ascending weight order: B1 B2 B3 B4 B5 B6 B7 B8
Bunnies before lunch: BBL
Left side of scale  : LFT
Right side of scale : RGT
(what tremendous imagination!)

WEIGH #1:
LFT = B1 + B2 + B7 + B8
RGT = B3 + B4 + B5 + B6

WEIGH #2:
if LFT = RGT then perform EVEN
if LFT > RGT then perform HEAVY
if LFT < RGT then perform LIGHT

EVEN
LFT = B3 + B4
RGT = B7
if LFT = RGT then BBL = 8
if LFT > RGT then BBL = 0
if LFT < RGT then BBL = 4

HEAVY
LFT = B1 + B4 + B8
RGT = B6 + B7
if LFT = RGT then BBL = 6
if LFT > RGT then BBL = 7
if LFT < RGT then BBL = 5

LIGHT
LFT = B2 + B3
RGT = B5
if LFT = RGT then BBL = 2
if LFT > RGT then BBL = 1
if LFT < RGT then BBL = 3
```

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