## Coin Change, Part 1

### May 17, 2013

It is a classic problem from computer science to determine the number of ways in which coins can be combined to form a particular target; as an example, there are 31 ways to form 40 cents from an unlimited supply of pennies (1 cent), nickels (5 cents), dimes (10 cents) and quarters (25 cents), ranging from the 3-coin set (5 10 25) to the 40-coin set consisting only of pennies.

The solution is usually stated in recursive form: if c is the first coin in the set of coins cs and n is the target, the solution is the number of ways to reach the target after removing c from cs plus the number of ways to reach nc using all the coins in cs. The algorithm to make a list of the coins, instead of the count, is the same, but keeping track of the list of coins instead of the count.

Your task is to write two functions, one to determine the count and one to determine the list of coins. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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### 7 Responses to “Coin Change, Part 1”

1. […] today’s Programming Praxis exercise, our goal is to list all of the ways in which a target amount can be […]

2. My Haskell solution (see http://bonsaicode.wordpress.com/2013/05/17/programming-praxis-coin-change-part-1/ for a version with comments):

```import Data.List

coins :: (Num a, Ord a) => [a] -> a -> [[a]]
coins _  0 = [[]]
coins xs n = [c:r | (c:cs) <- tails xs, c <= n, r <- coins (c:cs) (n-c)]

count :: (Num a, Ord a) => [a] -> a -> Int
count xs = length . coins xs
```
3. Alan S. said

Here is a Ruby program with both functions

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 \$ ./coin-change1.rb For total = 40 There are 31 sets of coins Set \$.25 \$.10 \$.05 \$.01 1: 0 0 0 40 2: 0 0 1 35 3: 0 0 2 30 4: 0 0 3 25 5: 0 0 4 20 6: 0 0 5 15 7: 0 0 6 10 8: 0 0 7 5 9: 0 0 8 0 10: 0 1 0 30 11: 0 1 1 25 12: 0 1 2 20 13: 0 1 3 15 14: 0 1 4 10 15: 0 1 5 5 16: 0 1 6 0 17: 0 2 0 20 18: 0 2 1 15 19: 0 2 2 10 20: 0 2 3 5 21: 0 2 4 0 22: 0 3 0 10 23: 0 3 1 5 24: 0 3 2 0 25: 0 4 0 0 26: 1 0 0 15 27: 1 0 1 10 28: 1 0 2 5 29: 1 0 3 0 30: 1 1 0 5 31: 1 1 1 0

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 #!/usr/bin/env ruby # coin-change1.rb # # count sets of coins (pennies, nickles, dimes, quarters) that # add to a specific total # # Count the different sets of the polynomial: # total = P + N*5 + D*10 * Q*25 # # A "coin set" is a 4-tuple of [P,N,D,Q] # \$coin_values = [ 1, 5, 10, 25 ] # given a total, find the maximum number of coins for each # denomination def compute_max_coins(total) \$max_coins = \$coin_values.map {|value| Integer(total / value)} \$max_pennies = \$max_coins[0] \$max_nickles = \$max_coins[1] \$max_dimes = \$max_coins[2] \$max_quarters = \$max_coins[3] end # get all the coin sets for a given total def get_coin_sets(total) coin_sets = [] (0..\$max_quarters).each { |q| break if total < q*25 qtotal = total – q*25 (0..\$max_dimes).each { |d| break if qtotal < d*10 dtotal = qtotal – d*10 (0..\$max_nickles).each { |n| break if dtotal < n*5 p = dtotal – n*5 coin_sets += [[q,d,n,p]] } } } coin_sets end if ARGV.size > 0 total = ARGV.shift.to_i else total = 40 end compute_max_coins(total) coin_sets = get_coin_sets(total) puts "For total = #{total}" puts "There are #{coin_sets.size} sets of coins" puts '' set = 1 printf "Set \$.25 \$.10 \$.05 \$.01\n" coin_sets.each do |q,d,n,p| printf "%3d: %4d %4d %4d %4d\n", set, q, d, n, p set += 1 end exit

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coin-change1.rb

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```#!/usr/bin/env ruby
# coin-change1.rb
#
# count sets of coins (pennies, nickles, dimes, quarters) that
# add to a specific total
#
# Count the different sets of the polynomial:
#   total = P + N*5 + D*10 * Q*25
#
# A "coin set" is a 4-tuple of [P,N,D,Q]
#

\$coin_values = [ 1, 5, 10, 25 ]

# given a total, find the maximum number of coins for each
# denomination

def compute_max_coins(total)
\$max_coins = \$coin_values.map {|value| Integer(total / value)}
\$max_pennies  = \$max_coins[0]
\$max_nickles  = \$max_coins[1]
\$max_dimes    = \$max_coins[2]
\$max_quarters = \$max_coins[3]
end

# get all the coin sets for a given total

def get_coin_sets(total)
coin_sets = []
(0..\$max_quarters).each { |q|
break if total < q*25
qtotal = total - q*25
(0..\$max_dimes).each { |d|
break if qtotal < d*10
dtotal = qtotal - d*10
(0..\$max_nickles).each { |n|
break if dtotal < n*5
p = dtotal - n*5
coin_sets += [[q,d,n,p]]
}
}
}
coin_sets
end

if ARGV.size > 0
total = ARGV.shift.to_i
else
total = 40
end

compute_max_coins(total)
coin_sets = get_coin_sets(total)

puts "For total = #{total}"
puts "There are #{coin_sets.size} sets of coins"
puts ''
set = 1
printf "Set  \$.25  \$.10  \$.05  \$.01\n"
coin_sets.each do |q,d,n,p|
printf "%3d: %4d  %4d  %4d  %4d\n", set, q, d, n, p
set += 1
end

exit
```
4. aks said

Here is the output:

```\$ ./coin-change1.rb
For total = 40
There are 31 sets of coins

Set  \$.25  \$.10  \$.05  \$.01
1:    0     0     0    40
2:    0     0     1    35
3:    0     0     2    30
4:    0     0     3    25
5:    0     0     4    20
6:    0     0     5    15
7:    0     0     6    10
8:    0     0     7     5
9:    0     0     8     0
10:    0     1     0    30
11:    0     1     1    25
12:    0     1     2    20
13:    0     1     3    15
14:    0     1     4    10
15:    0     1     5     5
16:    0     1     6     0
17:    0     2     0    20
18:    0     2     1    15
19:    0     2     2    10
20:    0     2     3     5
21:    0     2     4     0
22:    0     3     0    10
23:    0     3     1     5
24:    0     3     2     0
25:    0     4     0     0
26:    1     0     0    15
27:    1     0     1    10
28:    1     0     2     5
29:    1     0     3     0
30:    1     1     0     5
31:    1     1     1     0
```
5. […] This post presents C# solutions to a coin change problem as described inÂ https://programmingpraxis.com/2013/05/17/coin-change-part-1. […]

6. taoufik belaidi said

I have done this exercise using a simple method: I divide the total (T) by the first coin (C1). I take the integer part of this division (n1=T//c1). n1 represents the maximum number of possiblities of having c1 in any combinaison. After that, I do the same process for finding the total minus 0, 1,…, n1*c1 now using just the new set (C2, C3, C4).

Here is the program in Python, and it works well: the result is outputed very fast.

c1=25
c2=10
c3=29
c4=3

T=40

C=[]
for i in range(0, (T//c1)+1):
if T-(ic1)==0:
C.append([i,0,0,0])
continue
for j in range(0, ((T-i
c1)//c2)+1):
if T-ic1-jc2==0:
C.append([i,j,0,0])
continue
for w in range(0, ((T-ic1-jc2)//c3)+1):
if (T-ic1-jc2-wc3)%c4 == 0:
C.append([i,j,w,(T-i
c1-jc2-wc3)//c4])

for i in C:
print(i)
print(len(C))

7. […] solved the standard coin change problem in two previous exercises. The particular problem given here is to find the minumum number of coins that can be […]