An Integer Decomposition Problem

October 12, 2010

tags: Algorithm, Integer Decomposition, Parliament

There are a lot of integer decomposition problems. Here is one of them: decompose an integer N into k different integers so that the sum of them is N and the multiplication of them is maximized. A vivid version of this problem described below.

Parliament (source: Northeastern Europe 1998): New convocation of The Fool Land’s Parliament consists of ${N}$ delegates. According to the present regulation delegates should be divided into disjoint groups of different sizes and every day each group has to send one delegate to the conciliatory committee. The composition of the conciliatory committee should be different each day. The Parliament works only while this can be accomplished. You are to write a program that will determine how many delegates should contain each group in order for Parliament to work as long as possible.

Input: The input file contains a single integer ${N}$ ( ${5<=N<=1000}$ ).

Output: Write to the output file the sizes of groups that allow the Parliament to work for the maximal possible time. These sizes should be printed on a single line in ascending order and should be separated by spaces.

Sample Input

Sample Output

3 4

Before introduce the solution, let us review an inequality and the multiplication principle described below.

Theorem 1 (Inequality of Arithmetric and Geometric Means) For any list of ${n}$ nonnegative numbers ${x_1,x_2,\ldots,x_n}$ , we have $\displaystyle x_1x_2\ldots x_n {\leq \left(\frac{x_1+x_2,\ldots,x_n}{n}\right)}^n$

and the equality holds if and only if ${x_1=x_2=\ldots=x_n}$ .

Theorem 2 (Multiplication Principle) If a task consists of ${k}$ different operations ${o_1,o_2,\ldots,o_k}$ , and each operation ${o_i}$ can be done by ${m_i}$ ways. Then, there are in total ${m_1m_2\ldots m_k}$ different ways to complete the task.

Back to our problem on hand. Assume that all delegates are separated into ${k}$ disjoint groups, each of which has ${x_i}$ delegates. Then, according to multiplication principle, there are ${M=x_1x_2\ldots x_k}$ different ways to compose a conciliatory committee. According to the inequality of arithmetric and geometric means, ${M}$ reaches its maximal value when ${x_1=x_2=\ldots =x_k}$ .

Unfortunately, for any ${x_i}$ and ${x_j}$ , if ${i \neq j}$ , then ${x_i \neq x_j}$ , and ${x_i}$ should be an integer. However, the inequality of arithmetric and geometric means does give us some intuition: ${M}$ should be as larger as ${x_1,x_2,\ldots,x_k}$ get close to each other. The following lemmas confirms this intuition. Without loss of generality, assume ${x_1 <x_2 < \ldots < x_k}$ . Define the ${gap}$ of two integers ${x_1}$ and ${x_2}$ as the number of integers between them. Denote it by ${gap(x_1,x_2)}$ . For example, ${gap(5,8)=2}$ since there are 6 and 7 between them. If ${gap(x_1,x_2)=0}$ , then we say that ${x_1}$ and ${x_2}$ has ${no}$ gap.

Lemma 3 Let ${t=x_1+x_2}$ when ${t}$ , ${x_1}$ and ${x_2}$ are nonnegative integers and ${x_1 \neq x_2}$ . ${x_1x_2}$ has the maximal value if and only if ${gap(x_1,x_2) \leq 1}$ . Particularly, when ${t\geq 5}$ , the maximum of ${x_1x_2}$ is larger than ${t}$ strictly.

Given a increasing sequence of integers ${X_k = x_1,x_2,\ldots,x_k}$ , define the number of gaps in ${X_k}$ as the number of pairs ( ${x_i}$ , ${x_{i+1}}$ ) which ${gap(x_i,x_{i+1})>0}$ . For example, sequence ${1,2,5,6,8}$ has 2 gaps: one between 2 and 5, and the other between 6 and 8.

Lemma 4 ${M}$ reaches the maximal value if and only if there is at most one gap in ${X_k}$ and that gap is at most 1 if any.

The lemma above tells us that to make ${M}$ as large as possible, the sequence ${X_k}$ should composed by a list of continous integers ${x+1,x+2,\ldots,x+k}$ , or two segments of continous integers ${x+1,x+2,\ldots,x+u}$ and ${(x+u+1)+1,(x+u+1)+2,\ldots,(x+u+1)+v}$ where ${u+v = k}$ . Example: ${N=15}$ , ${k=3}$ . Then ${M=4\times 5\times 6=120}$ . If ${k=4}$ , then ${M=2\times 3\time 4\times 6=144}$ . For ${k=5}$ , ${M=1\times 2\times 3\times 4\times 5 = 120}$ .

One natural question is that, given ${k}$ is fixed, how to look for valid sequence ${X_k}$ which maximize ${M}$ ? Well, by lemmas above, we can assume that ${X_k= x+1,x+2,\ldots,x+u, (x+u+1)+1,(x+u+1)+2,\ldots,(x+u+1)+v}$ where ${u + v = k}$ and ${v < k}$ . So

$\displaystyle (\sum_{i=1}^u x+i) + (\sum_{j=1}^v (x+u+1)+j) = N$

$\displaystyle x=\frac{N}{k}-\frac{v}{k}-\frac{k+1}{2}$

by assuming ${N=mk+t}$ where ${t < k}$ ,

$\displaystyle x=m-\frac{t-v}{k}-\frac{k+1}{2}$

Since ${x}$ is integer and ${v<k}$ , when ${k}$ is odd, ${v=t}$ . When ${k}$ is even, ${v=t\pm \frac{k}{2}}$ , depending on if ${t}$ is larger than half of ${k}$ . By this way, for any ${N}$ and valid ${k}$ , we compute the unique pair of ${x}$ and ${v}$ .

Lemma 5 Given ${N}$ and ${k}$ , the sequence ${X_k}$ maximizing ${M}$ is unique.

At this point, we can solve the problem as follows: For all possible ${k}$ , compute the ${x}$ and ${v}$ by above way. Then compute their corresponding ${M}$ values, between which we pick the largest one. The algorithm is inefficient since we involve computing the pretty number of many factors (imagine ${4\times5\times\ldots\times 50}$ !!). Java and some programming language do provide BigInteger class, but it is time consuming.

The question following is, can we void manipulation on big integers? Yes. To make it, we need some insightful thoughts. From the example given above, it seems that ${M}$ increases and ${x_1}$ decreases when ${k}$ get larger, but not too large. At least, ${x_1 = 1}$ is not expected and should be voided. In a word, we try to get an ${x_1\geq 2}$ as small as possible. The perfect value for ${x_1}$ would be 2, of course. However, not always we can make it. The good news is, ${x_1}$ should be 2 or 3.

Lemma 6 Let ${X_k}$ be the sequence maximinze ${M}$ among all valid sequences. Then ${x_1}$ is 2 or 3.

Proof: We prove it by contradiction. Assume ${x_1 > 3}$ . Consider the case ${x_1=4}$ . By Lemma 4, ${x_2=5}$ or ${6}$ . If ${x_2=5}$ , we construct a new sequence ${X^\prime}$ by replacing ${x_2}$ with ${x_1^\prime=2}$ and ${x_2^\prime=3}$ and keeping others intacted. Note that ${X^\prime}$ is a valid sequence, meaning that all elements in it are different and sum to ${N}$ . Then ${\frac{X_k}{X^\prime} = \frac{5}{6}}$ . That’s ${X_k < X^\prime}$ , which is a contradiction. If ${x_2=6}$ , we construct a new valid ${X^\prime}$ by replacing ${x_1}$ and ${x_2}$ by 2,3 and 5. Since ${2\times 3\times 5 = 30 > 4\times 6=24}$ , it contradicts the optimality of ${X_k}$ . Consider ${x_1 > 4}$ . In this case, we construct the new valid ${X^\prime}$ by replacing ${x_1}$ with two small factors ${y_1}$ and ${y_2}$ such that ${gap(y_1,y_2)\leq 1}$ . According to Lemma 5, ${X^\prime > X_k}$ , yet another contradiction. So ${x_1 \leq 3}$ . Obviously, if ${x_1=1}$ , we can easily construct an valid ${X^\prime > X_k}$ .
Now, we narrow the search scope to those ${X_k}$ starting with 2 or 3. Furthermore,

Lemma 7 Let ${X=x+1,x+2,\ldots,x+u_1, (x+u_1+1)+1,\ldots,(x+u_1+1)+v_1}$ be the sequence maximizing ${M}$ starting from 2 ( ${x=1}$ ) and ${Y=y+1,y+2,\ldots,y+u_1,(y+u_2+1)+1,\ldots,(y+u_2+1)+v_2}$ starting from 3 ( ${y=2}$ ), where ${u_1,u_2 \geq 1}$ and ${v_1,v_2 \geq 0}$ . Let ${i}$ be the length of ${X}$ and ${j}$ be the length of ${Y}$ . Then, ${i>j}$ and ${v_1 < v_2}$ .

Proof: Obviously, ${v_1 < i}$ . ${\sum_{a\in X} a = N \Rightarrow (2i+3)^2=8N-8v_1+9}$ and ${\sum_{a\in Y} a = N \Rightarrow (2j+5)^2=8N-8v_2+25}$ . Combining them, we have $\displaystyle (j-i+1)(i+j+4) = 2(v_1-v_2)+4$

Assume ${i\leq j}$ . Then ${2(v_1-v_2)+4 \geq i+j+4 \geq 2i+4}$ , or ${v_1-v_2 \geq i}$ . However, since ${v_1 < i}$ and ${v_2 \geq 0}$ , it’s impossible. So ${i > j}$ and then ${v_1 < v_2}$ .

The above lemma simply states that if we get two valid sequences, namely, one starting from 2 with length ${i}$ which maximize ${M}$ when ${k=i}$ , and the other starting from 3 with length ${j}$ which maximize ${M}$ when ${k=j}$ , and each of them consists of two segments of continguous integers, then latter segment of the one starting from 2 is short than that of the one starting from 3.

This fact leads to following important lemma.

Lemma 8 Let valid ${X=x+1,x+2,\ldots,x+u,(x+u+1)+1,\ldots,(x+u+1)+v}$ start from 3 ( ${x=2}$ ). Then ${X}$ maximizes ${M}$ among all valid sequences if and only if ${v=0}$ or ${v=1}$ .

Proof: In either case of ${v=0}$ or ${v=1}$ , no valid sequence starting from 2 can be constructed according Lemma 7. According to Lemma 6, ${X}$ maximizes ${M}$ . Now assume ${X}$ maximize ${M}$ but ${v>1}$ . Then, we construct a new valid ${Y}$ by replacing ${(x+u+1)+2}$ with 2 and ${(x+u+1)}$ . Since ${u>0}$ and ${x=2}$ , ${2(x+u+1) > (x+u+1)+2}$ . So ${Y>X}$ . Contradiction.

Now we come to the crux of the problem. We construct a valid sequence ${X}$ starting from ${x+1}$ where ${x=1}$ by the following way: start from ${x+1}$ , we keep adding ${x+2,\ldots,x+i}$ until adding ${x+i+1}$ will make the sum of elements in ${X}$ larger than ${N}$ . Then, we increases each element by 1 in the order of ${x+i,x+i-1,\ldots,x+1}$ and repeat the process until it sums to ${N}$ . The sequence we finally get is the answer.

${S \gets 0}$
${k \gets 0}$
array ${X}$ stores the sequence
while ${S + (k+2) \leq N}$
       ${X[k+1] \gets k+2}$
       ${k\gets k+1}$
       ${S \gets S + X[k]}$
end while
${j \gets k}$
${m \gets N-S}$
while ${m > 0}$
       ${X[j] \gets X[j]+1}$
       ${m \gets m-1}$
       ${j \gets j-1}$
      if ${j < 1}$
             ${j = k}$
      end if
end while

The next step is to prove the algorithm is correct. Given above lemmas, it’s an easy task.

Proof: Firstly, ${m \leq k+1}$ . Otherwise, the first while loop won’t terminate. So the body of second while loop will be executed at most ${k+1}$ times. If ${m=k}$ or ${m=k+1}$ , then ${X}$ starts from 3 with the second segment of length 0 or 1, respectively. According to Lemma 8, ${X}$ is optimal. If ${m = 0}$ , we are unable to construct a sequence ${Y}$ such that ${Y}$ starts from 3 and will the second segment of length at most 1. If ${0<m<k}$ , ${X}$ has second segment of length at least 1. According to Lemma 7, we also cann’t construct ${Y}$ maximing ${M}$ and starting from 3 with second segment of length at most 1. Combining Lemma 6, the ${X}$ is optimal.