IQ / Programming Quiz (Cannon-Ball Stacks)

[Lost User]

The way I see it, each stack must have a number of cannonballs from the set {1, 4, 10, 20, 35...} Now we are looking for a minimum A and B, both out of this set, so that the sum of both is also in the set. From your example we know it works for two stacks of 10 cannonballs, so the sum cannot be greater than 20. Therefore A and B must both be in {1, 4, 10} Now you can simply try all combinations:

A | B | Sum | Sum <= 20 | Sum in set

1 | 1 | 2 | Y | N
1 | 4 | 5 | Y | N
1 |10 | 11 | Y | N
4 | 1 | 5 | Y | N
4 | 4 | 8 | Y | N
4 |10 | 14 | Y | N
10| 1 | 11 | Y | N
10| 4 | 14 | Y | N
10|10 | 20 | Y | Y

So here we have it. Your example with two stacks of 10 cannonballs already is the minimum. Edit: Added pre tags to make the table readable

I'm invincible, I can't be vinced

[BillWoodruff]

Hi AspDotNetDev, In this comment to your good self(ves?)[^], I think you raised an issue appropriate for "Site Suggs and Buggs:" and I have expressed my views there:[^]. And, to this starving worm on a wilted cabbage leaf, the issue raised by that post is an even more compelling puzzle than the stacking of cannon-balls, although I am not denigrating the value of considering the stacking of cannon balls ! best, Bill

"Science is facts; just as houses are made of stones: so, is science made of facts. But, a pile of stones is not a house, and a collection of facts is not, necessarily, science." Henri Poincare

[AspDotNetDev]

Nope, there was a stipulation that A and B are different. Otherwise, you are one the right track.

Thou mewling ill-breeding pignut!

[AspDotNetDev]

leppie wrote:

Can you not see the symmetry?

They both involve triangles. Other than that, nope.

Thou mewling ill-breeding pignut!

[Thor Sigurdsson]

A=20, B=54, C=55 or B=54, A=20, C=55 used two recurisve functions and iteration (lazy mans search).

I=I.am()?Code(I):0/0;

[YvesDaoust]

1. looked up the tetrahedral numbers in Wikipedia and found this list: 1, 4, 10, 20, 35, 56, 84, 120, 165, 220, 286, 364, 455, 560, 680, 816, 969... (10 seconds of work) 2) opened an Excel sheet and created the table of sums 2 5 11 21 36 57... 5 8 14 24 39 60... 11 14 20 30 45 66... 21 24 30 40 55 76... 36 39 45 55 70 91... ... (just needed to recall the appropriate syntax to form the cell expression; 1 minute of work) 3) sorted all sums in increasing order 2, 5, 5, 8, 11, 11, 14, 14, 20, 21, 21, 24, 24, 30, 30, 36, 36, 39, 39, 40, 45, 45, 55, 55, 57, 57, 60, 60, 66, 66, 70, 76, 76, 85, 85, 88, 88, 91, 91, ... (moved to Word to flatten the table structure; 1 minute of work) 4) spotted by eye in the list the first tetrahedral number larger than 20: 680, the sum of 120 and 560. (Lucky the Wikipedia list was long enough :)) (2 extra minutes) 5) explained the answer in CodeProject (half an hour)

[jsc42]

I did this using an old technique: pencil and paper. It is more complicated to explain than it was to do. Basically, I wrote 3 columns of numbers: Col 1: 1, 2, 3, 4, etc [represents the no of levels in the pyramid) Col 2: 1, (corresponding no in col 1 + prev number in col 2), ... [represents the no of new balls in the new level of the pyramid] Col 3: 1, (corresponding no in col 2 + prev number in col 3), ... [represents the total no of balls in the pyramid] and then looked for a number in col 2 that was the same as a number in col 3 which is a pyramid's worth of balls that is also a level's worth of balls [Actually, I didn't bother with col 1 - I could work that one out without writing it down, but it is easier to explain with it there] My solution is ... The first one found was 120, so the solution is 120 + 560 (no in col 3 before the 120 in col 2) = 680 (no in col 3 next to 120 in col 2). (select the text in the gap above, e.g. by dragging the mouse, to read it) Of course, this discounts the possibility that a pyramid might split across multiple levels in a combined pyramid. Just in case I had made a simple arithmetic error, I checked in MS-Excel: Cell A1 = 1, B1 = 1, C1 = 1 Cell A2 = =A1+1, B2 = =A2+B1, C2 = =B2+C1 Cells A3 through C17 = Copy and paste A2 through C2 [I've had to edit this entry twice - suffering from lysdexia (!) today].

[yiangos]

Well, using a bit of python:

def pyr(l):
if l==0:
return 0
elif l==1:
return 1
else:
return pyr(l-1)+l*(l+1)/2

#then we brute-force
a=[pyr(i) for i in range(100)][1:]
b=[(i,j,k) for i in a
for j in a
for k in a
if (i!=j and i+j==k)]
if len(b)>0:
print b[0]

Running the above yields the result as

(120, 560, 680)

as others have already pointed out... (Edit: changed the type from general to Answer)

Φευ! Εδόμεθα υπό ρηννοσχήμων λύκων! (Alas! We're devoured by lamb-guised wolves!)

[YvesDaoust]

A Python script can do. Grow a list of tetrahedral numers; for every new number, try and find a number in the list such that when subtracted from the new number, it gives another number in the list. (This way you make sure that for every new number the two terms of the decomposition are already in the list.)

# Start with an empty list
List= {}
n= 1

while n > 0:
# Try the next tetrahedral number
r= n * (n + 1) * (n + 2) / 6

# Try every number in the list
for p in List:
    # Lookup the difference between the new number and the current one
    q= r - p
    if p != q and q in List:
        print p, '+', q, '=', r

        # Flag as found
        n= -1
        break

# Not found, extend the list
List\[r\]= None
n+= 1

For efficiency of the search, the list is implemented as a dictionnary. Yields:

560 + 120 = 680

[el_vez]

Haskell answer:

layers = scanl1 (+) [1..]
sizes = scanl1 (+) layers
combinedSizes = [(s1, s2, s1 + s2) | s1 <- sizes, s2 <- takeWhile (< s1) sizes]
newPyramids = filter combinesToPyramid combinedSizes
where combinesToPyramid (s1, s2, sum) = sum `elem` (takeWhile (sum >=) sizes)

answer = head newPyramids

-------- SPOILER ------------ this gives the answer as "(560, 120, 680)", (if you load it up in GHCI and type "answer"). If you want the n first solutions, just type "take n newPyramids":

take 3 newPyramids
[(560,120,680),(27720,1540,29260),(29260,4960,34220)]

[YvesDaoust]

Brute force:

def T(n):
return n * (n + 1) * (n + 2) / 6

r= 1
while True:
q= 1
while q < r:
p= 1
while T(p) + T(q) < T(r):
p+= 1
if T(p) + T(q) == T(r):
print T(p), '+', T(q), '=', T(r)
q+= 1
r+= 1

Yields:

10 + 10 = 20
560 + 120 = 680
120 + 560 = 680
27720 + 1540 = 29260
1540 + 27720 = 29260
29260 + 4960 = 34220
4960 + 29260 = 34220
59640 + 10660 = 70300
10660 + 59640 = 70300
182104 + 39711 = 221815
39711 + 182104 = 221815
...

[YvesDaoust]

Even bruter force:

def T(n):
return n * (n + 1) * (n + 2) / 6

for r in range (1, 1000):
for q in range(1, r):
for p in range(1, q):
if T(p) + T(q) == T(r):
print T(p), '+', T(q), '=', T(r)

Yields:

120 + 560 = 680
1540 + 27720 = 29260
4960 + 29260 = 34220
10660 + 59640 = 70300
39711 + 182104 = 221815
102340 + 125580 = 227920
7140 + 280840 = 287980
19600 + 447580 = 467180
...

[PhilLenoir]

Sorry, 6 on the bottom, 3 next then 1 = total 10 for a 3x3x3 pyramid!

Life is like a s**t sandwich; the more bread you have, the less s**t you eat.

[AspDotNetDev]

Nope, 20 + 54 is not 55.

Thou mewling ill-breeding pignut!

[PhilLenoir]

Maybe not the most elegant solution but might possibly be the fastest way to come up with the answer!: 1 Using Excel I created 3 columns: column 1 being a linear sequence (1,2,3 ....100); column 2 representing each layer (B1=1,B2=A1+A2); the third representing the pyramid (C1=1,C2=B1+B2) 2 I pasted these into an Access table 3 I cross joined the table to itself and added the two columns 3 4 I inner joined the query to the table on sum = column 3 Answer 120 (8 layers) + 560 (14 layers) = 680 (15 layers)

Life is like a s**t sandwich; the more bread you have, the less s**t you eat.

[RolfReden]

got one, just by accident :) 560 + 120 = 680, which are an 8-layered and a 14-layered pyramid combined to a 15-layered pyramid. still thinking of an smart search or algebraic solution tho [edit for some code] clear clc lines(0) n = 30 m(1) = 1 k(1) = 1 for i = 1:n m(i+1) = m(i) + 1 + (i) k(i+1) = k(i) + m(i+1) end [/edit]

[Stephen Dycus]

Wow such complicated answers.... the answer is 4. The question is If the two smaller pyramids are different sizes, however, what would be the minimum number of cannon-balls he could use to make one large tetrahedral pyramid? ... but the thing is, there are still only 20 cannonballs to use to build the pyramid and he never said you had to use all the cannonballs. It asks the MINIMUM... which is 3 on the bottom + 1 on top.

[PhilLenoir]

How do 3 cannonballs make a tetrahedral pyramid? :P

Life is like a s**t sandwich; the more bread you have, the less s**t you eat.

[Stephen Dycus]

I didn't say 3 I said 4 :P

[PhilLenoir]

... but the source has to be two tetrahedral pyramids ...

Life is like a s**t sandwich; the more bread you have, the less s**t you eat.