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hello-algo/en/codes/ruby/chapter_dynamic_programming/knapsack.rb
Yudong Jin 14c314c08c Retranslate the complete English edition from Simplified Chinese (#2010)
* Retranslate the complete English edition from Simplified Chinese

* Polish three English explanations after reader-focused review
2026-10-31 03:35:51 +08:00

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Ruby

=begin
File: knapsack.rb
Created Time: 2024-05-29
Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
=end
### 0-1 knapsack: brute-force search ###
def knapsack_dfs(wgt, val, i, c)
# If all items have been considered or no capacity remains, return value 0
return 0 if i == 0 || c == 0
# If the item exceeds the remaining capacity, it must be excluded
return knapsack_dfs(wgt, val, i - 1, c) if wgt[i - 1] > c
# Calculate the maximum values obtained by excluding and including item i
no = knapsack_dfs(wgt, val, i - 1, c)
yes = knapsack_dfs(wgt, val, i - 1, c - wgt[i - 1]) + val[i - 1]
# Return the larger value from the two choices
[no, yes].max
end
### 0-1 knapsack: memoized search ###
def knapsack_dfs_mem(wgt, val, mem, i, c)
# If all items have been considered or no capacity remains, return value 0
return 0 if i == 0 || c == 0
# If a result has already been recorded, return it directly
return mem[i][c] if mem[i][c] != -1
# If the item exceeds the remaining capacity, it must be excluded
return knapsack_dfs_mem(wgt, val, mem, i - 1, c) if wgt[i - 1] > c
# Calculate the maximum values obtained by excluding and including item i
no = knapsack_dfs_mem(wgt, val, mem, i - 1, c)
yes = knapsack_dfs_mem(wgt, val, mem, i - 1, c - wgt[i - 1]) + val[i - 1]
# Record and return the larger value from the two choices
mem[i][c] = [no, yes].max
end
### 0-1 knapsack: dynamic programming ###
def knapsack_dp(wgt, val, cap)
n = wgt.length
# Initialize the dp table
dp = Array.new(n + 1) { Array.new(cap + 1, 0) }
# State transitions
for i in 1...(n + 1)
for c in 1...(cap + 1)
if wgt[i - 1] > c
# If item i exceeds the remaining capacity, do not choose it
dp[i][c] = dp[i - 1][c]
else
# Take the larger value obtained by excluding and including item i
dp[i][c] = [dp[i - 1][c], dp[i - 1][c - wgt[i - 1]] + val[i - 1]].max
end
end
end
dp[n][cap]
end
### 0-1 knapsack: space-optimized dynamic programming ###
def knapsack_dp_comp(wgt, val, cap)
n = wgt.length
# Initialize the dp table
dp = Array.new(cap + 1, 0)
# State transitions
for i in 1...(n + 1)
# Traverse in decreasing order
for c in cap.downto(1)
if wgt[i - 1] > c
# If item i exceeds the remaining capacity, do not choose it
dp[c] = dp[c]
else
# Take the larger value obtained by excluding and including item i
dp[c] = [dp[c], dp[c - wgt[i - 1]] + val[i - 1]].max
end
end
end
dp[cap]
end
### Driver code ###
if __FILE__ == $0
wgt = [10, 20, 30, 40, 50]
val = [50, 120, 150, 210, 240]
cap = 50
n = wgt.length
# Brute-force search
res = knapsack_dfs(wgt, val, n, cap)
puts "Maximum total item value within the knapsack capacity: #{res}"
# Memoized search
mem = Array.new(n + 1) { Array.new(cap + 1, -1) }
res = knapsack_dfs_mem(wgt, val, mem, n, cap)
puts "Maximum total item value within the knapsack capacity: #{res}"
# Dynamic programming
res = knapsack_dp(wgt, val, cap)
puts "Maximum total item value within the knapsack capacity: #{res}"
# Space-optimized dynamic programming
res = knapsack_dp_comp(wgt, val, cap)
puts "Maximum total item value within the knapsack capacity: #{res}"
end