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How many ways can you distribute <i>n</i> different objects into <i>k</i>
nonempty bundles? Traditionally this number is denoted by
$\displaystyle\left\{{n \atop k}\right\},$ for example $\displaystyle\left\{{4 \atop 2}\right\}=7$ because 4
objects — call them A, B, C, D — can be distributed
into two bundles
in the seven ways
<div style='text-align:center'>A-BCD B-ACD C-ABD D-ABC AB-CD AC-BD AD-BC</div>
To determine the value of these numbers, we will use the recurrence relation
<ul>
<li>for $\displaystyle n,k>0, \left\{{n \atop k}\right\}
= k\left\{{n-1 \atop k}\right\}+\left\{{n-1 \atop k-1}\right\}$</li>
<li>with the base cases $\displaystyle\left\{{0\atop 0}\right\}=1$ and
for $\displaystyle n>0,
\left\{{n \atop 0}\right\}=0$ and for
$\displaystyle k>0, \left\{{0 \atop k}\right\}=0$
</li>
</ul>
Write a static method <code>bundleWays(n, k)</code> that uses
dynamic programming to compute $\displaystyle\left\{{n \atop k}\right\}.$
Using a recursive method will be too slow.

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Correct answer
Template / Reference solution

public static long bundleWays(int n, int k) {
\[
long[][] dp = new long[n+1][k+1];

// base cases are automatically set to zero, except this one:
dp[0][0] = 1;

// compute numbers from bottom-up
for (int i=1; i<=n; i++) {
for (int j=1; j<=k; j++) {
dp[i][j] = j*dp[i-1][j] + dp[i-1][j-1];
}
}

return dp[n][k];
]\
}

public static void main(String[] args) {
for (int n=0; n<=5; n++) {
for (int k=0; k<=n; k++)
StdOut.printf("{%d %d} = %-3d ", n, k, bundleWays(n, k));
StdOut.println();
}
}

Template / Reference solution

Java test suite
See manual

test("bundleWays", 4, 2);
testMain();
test("bundleWays", 25, 2);
test("bundleWays", 25, 9);
test("bundleWays", 25, 10);
test("bundleWays", 25, 11);
test("bundleWays", 25, 20);

C++ test suite
json list of stdin/args tests
e.g. [{"stdin":"hi", "args":["4", "5"]}, {"stdin":"noargs"}]
to just run once with no input use [{}]

C++ test suite
See manual

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Property	Value
Description html markup shown to student	How many ways can you distribute <i>n</i> different objects into <i>k</i> nonempty bundles? Traditionally this number is denoted by $\displaystyle\left\{{n \atop k}\right\},$ for example $\displaystyle\left\{{4 \atop 2}\right\}=7$ because 4 objects — call them A, B, C, D — can be distributed into two bundles in the seven ways <div style='text-align:center'>A-BCD B-ACD C-ABD D-ABC AB-CD AC-BD AD-BC</div> To determine the value of these numbers, we will use the recurrence relation <ul> <li>for $\displaystyle n,k>0, \left\{{n \atop k}\right\} = k\left\{{n-1 \atop k}\right\}+\left\{{n-1 \atop k-1}\right\}$</li> <li>with the base cases $\displaystyle\left\{{0\atop 0}\right\}=1$ and for $\displaystyle n>0, \left\{{n \atop 0}\right\}=0$ and for $\displaystyle k>0, \left\{{0 \atop k}\right\}=0$ </li> </ul> Write a static method <code>bundleWays(n, k)</code> that uses dynamic programming to compute $\displaystyle\left\{{n \atop k}\right\}.$ Using a recursive method will be too slow.
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Remarks Comments, history, license, etc.	remove
Engine
Choices json list of string, bool pairs e.g. `[["good", true], ["bad", false]]`
Correct answer
Template / Reference solution	public static long bundleWays(int n, int k) { \[ long[][] dp = new long[n+1][k+1]; // base cases are automatically set to zero, except this one: dp[0][0] = 1; // compute numbers from bottom-up for (int i=1; i<=n; i++) { for (int j=1; j<=k; j++) { dp[i][j] = j*dp[i-1][j] + dp[i-1][j-1]; } } return dp[n][k]; ]\ } public static void main(String[] args) { for (int n=0; n<=5; n++) { for (int k=0; k<=n; k++) StdOut.printf("{%d %d} = %-3d ", n, k, bundleWays(n, k)); StdOut.println(); } }
Template / Reference solution
Java test suite See manual	test("bundleWays", 4, 2); testMain(); test("bundleWays", 25, 2); test("bundleWays", 25, 9); test("bundleWays", 25, 10); test("bundleWays", 25, 11); test("bundleWays", 25, 20);
C++ test suite json list of stdin/args tests e.g. `[{"stdin":"hi", "args":["4", "5"]}, {"stdin":"noargs"}]` to just run once with no input use `[{}]`
C++ test suite See manual
Forbidden substrings json list of strings e.g. `["for","while"]`	remove
Solution visibility	remove
Java imports json list of importables e.g. `["java.util.*"]`	remove
Override Java classname else defaults to websheet name	remove
Dependent on other websheets? json list of websheet names in this folder e.g. `["DataStructure"]`	remove
Is example? i.e., just a demo	remove
Add compiler flag(s) json list of flags e.g. `["-Wno-unused-variable"]`	remove
Remove default compiler flag(s) json list of flags see default_cppflags e.g. `["-Wno-write-strings"]`	remove

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