# Self-counting sentences II

Let’s revisit the topic of the previous post on self-counting sentences. We looked at sentences like this:

In this sentence the number of occurrences of 1 is __, of 2 is __, ..., of n is __.

Let’s go up one more step on the ladder of abstraction, to arrive at sentences of the type:

In this sentence, there are 3 numbers appearing 1 time, 1 number appearing 2 times, 1 number appearing 3 times, 0 numbers appearing 4 times.

In this sentence, there are 2 numbers appearing 1 time, 3 numbers appearing 2 times, 0 numbers appearing 3 times, 0 numbers appearing 4 time.

Observe, that both say the truth about themselves. Below are three puzzles. In all cases the task is to fill in the gaps such as to make the sentence say the truth and the questions are: for what values of n does a solution exist and what are the solutions (if they exist).

In this sentence there are __ numbers occurring 1 time, __ numbers occurring 2 times, ..., __ numbers occurring n times.

In this sentence, from the numbers 0,1, ..., n-1, there are __ numbers occurring underlined 0 time, __ numbers occurring underlined 1 time, ..., __ numbers occurring underlined n-1 times.

In this sentence there are __ numbers occurring underlined 1 time, __ numbers occurring underlined 2 times, ..., __ numbers occurring underlined n times.

Note: I submitted the first variant of the problem to Kömal, where it appeared in the 2013 September issue (Hungarian, English – unfortunately, as of this writing, the English version appears mistaken there). Later, Toshihiro Shimizu pointed out that a formulation essentially equivalent to the second variant above appeared earlier as an IMO problem (Combinatorics 5)

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The second problem is equivalent to the first problem. There exists a solution to problem 1, if and only if the same sequence of numbers is a solution to problem 2 (note that in problem 2 the counting starts from 0). Therefore we only look at the first and the third problems.

Solution of problem 1:

For small values of n, we can find the solutions by trial and error or with a small program. We denote the numbers of the solution as f(1), f(2), …, f(n).

For n=1, there is a single solution: f(1)=2.

For n=2, there is a single solution: f(1)=4, f(2)=0.

For n=3, there is a single solution: f(1)=4, f(2)=1, f(3)=0.

For n=4, there are two solutions (see the beginning of the post).

For n=5, n=6, there are no solutions.

For n=7, there are two solutions: f(1)=4, f(2)=3, f(3)=0, f(4)=1, f(5)=0, f(6)=0, f(7)=0, and f(1)=5, f(2)=1, f(3)=1, f(4)=1, f(5)=0, f(6)=0, f(7)=0.

For n=8, there are two solutions: f(1)=5, f(2)=2, f(3)=1, f(4)=1, f(5)=0, f(6)=0, f(7)=0, f(8)=0, and f(1)=5, f(2)=3, f(3)=0, f(4)=0, f(5)=1, f(6)=0, f(7)=0, f(8)=0.

For every n>=9 there are exactly three solutions:
a) f(1)=n-3, f(2)=3, f(n-3)=1, and the remaining values 0,
b) f(1)=n-2, f(2)=1, f(4)=1, f(n-4)=1, and the remaining values 0,
c) f(1)=n-3, f(2)=2, f(3)=1, f(n-4)=1, and the remaining values 0.

These solutions clearly work. Let’s prove that no other solutions exist.

We can observe two properties of the sequence f(i):

Property 1: ∑ i * f(i) = 2n. (where i goes from 1 to n)
Proof:
There are 2n numbers in the sentence and each i*f(i) term counts how many numbers are there with multiplicity i. To have a multiplicity greater than n would mean that all gaps are filled with the same value, which cannot yield a true sentence for any value. Thus the possible multiplicities are between 1 and n, therefore ∑ i * f(i) gives the total number of numbers in the sentence, which is 2n.

Property 2: ∑ f(i) = n+1. (where i goes from 1 to n)
Proof:
Clearly, for all i, f(i)>=0 and f(i)<=n. If for all i, f(i)>0 would hold, ∑ i * f(i) would be larger than 2n, violating Property 1. Therefore, some of the f(i) values have to be equal to 0, which means that all numbers 0,1,…,n appear in the sentence. Furthermore, all numbers that appear in the sentence have multiplicity between 1 and n. Thus, ∑ f(i) gives the total number of distinct values, which is n+1.

Now we prove that a) b) c) are the only solutions for n>=9.

If f(1)>n-2,
it must be that f(1)=n-1, but now n-1 occurs twice, so to maintain that n-1 values occur once, we need to write n-1 in every other position as well. This is not a correct solution.

If f(1)=n-2,
we must have exactly one other nonzero value (besides n-2) written in the gaps. We cannot write n-2 anywhere else, as that would violate Property 2. To get a sum of n+1, the possibilities are writing one time 3, or three times 1. Writing 3 anywhere doesn’t make the sentence correct, so it remains to write three 1’s. But then the number of 0’s will be n-4, the number of 1’s will be 4 and the number of (n-2)’s will be 2. This gives us solution b).

If f(1)=n-3,
we have two other nonzero values (besides n-3) written in the gaps. They also have to add up to 4 (Property 2). That can be as 3+1 or as 2+1+1. The first case means that 1,3 and n-3 have multiplicities 2 and 0 has multiplicity n-3. The second case means that 2 and n-3 have multiplicity 2, 0 has multiplicity n-4 and 1 has multiplicity 3. These cases uniquely lead to solutions a) and c).

If f(1)<n-3,
suppose f(1)=n-k (where k>3). The value 0 has to appear more than one time (otherwise the sum of Property 1 would be too large). Therefore, among the values 1, …, n, there are k distinct values which are written somewhere in the gaps (to make their multiplicities larger than one in the whole sentence). We know that n-k appears in the first gap, so we have k-1 other values in the gaps starting from the second. Let’s denote these values by a1 < a2 < . . .  < ak − 1. Since they are written in the gaps after the first one, it follows that there are at least a1 + a2 + . . .  + ak − 1 distinct numbers that appear more than once in the sentence. Even if one of them is n-k, and one of them is 0, that still leaves S = a1 + a2 + . . .  + ak − 1 − 2 distinct numbers appearing more than once (therefore, also in the gaps). Since a1 ≥ 1, we have S >= k(k-1)/2 – 2. Even if the smallest of these numbers is 1, we get that f(i) >> n+1, for k>3. This violates Property 2, therefore no solution of this kind is possible.

Solution of problem 3:

For n=1, we have f(1)=1, unique solution.
For n=2, we have f(1)=2, f(2)=0, unique solution.
For n=3, we have f(1)=1, f(2)=1, f(3)=0, unique solution.
For n=4, we have f(1)=2, f(2)=1, f(3)=0, f(4)=0, unique solution.

For n>=5, we have exactly two solutions:
a) f(2)=1, f(n-2)=1, the remaining values 0
b) f(1)=2, f(n-2)=1, the remaining values 0.

Proof:

Similarly to the first problem, now we have the property:
∑ i * f(i) = n.

Clearly f(n)=0, otherwise all values would have to be equal, which is not a correct solution.
Also, f(n-1)=0. The only alternative would be f(n-1)=1, and all other values equal to zero, but this is not a correct solution either.

If f(n-2)=1 we need that ∑ i * f(i) over all other values is 2. This leads to solutions a) and b) only.
f(n-2)>=2 is not possible, as it would violate the property for n>=5.

Now suppose f(n-k) is the last nonzero value, for k>=3. This means that there are k numbers that are not equal to one of the most frequent values. If 0 is one of the most frequent values, then there are k nonzero values. All of them cannot be equal to 1, as that would not be a correct solution, so ∑ i * f(i) >= k(k-1)/2 + n-k + 1, which is larger than n for k>=3, violating the required property. On the other hand, if less than half of the elements are 0, the property is clearly violated, which shows that there are no other solutions.

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