Followup to Product & Sum.

Contradicting what was said in the previous post, the *blue eyed islanders puzzle* will be solved in a future post to avoid making this post excessively long. 😀

#### Product & Sum

One way to visualize the structure of this problem is to take each assertion by P and S as a filter to be applied over the initial set of integer pairs. This solution will then represent each of these filters as a block of Python code, with additional text explaining how each filter is connected with the associated assertion.

Given are X and Y, two integers, greater than 1, are not equal, and their sum is less than 100. S and P are two mathematicians; S is given the sum X+Y, and P is given the product X*Y of these numbers.

It is clear that we can restrict the first number to be strictly less than the second number, as the order of the numbers cannot be determined from the given data. Then we can get all the allowed pairs with the following Python code:

all_pairs = [(x, y) for y in range(2, 100) for x in range(2, y) if x + y < 100]

– P says “I cannot find these numbers.”

This implies that there are multiple allowed pairs whose products match the value that was given to P. We can start counting the number of pairs with each possible product:

num_pairs_by_prod = {} for x, y in all_pairs: if x * y not in num_pairs_by_prod: num_pairs_by_prod[x * y] = 0 num_pairs_by_prod[x * y] += 1

The pairs allowed by P’s assertion are those whose product is shared with other pairs:

pairs_1 = [(x, y) for x, y in all_pairs if num_pairs_by_prod[x * y] > 1]

– S says “I was sure that you could not find them.”

Then we know that the value of the sum is enough to know that the product cannot identify the pair of integers. The set of sums of integer pairs that *can* be identified by their products is:

identif_by_prod_pairs_sums = set(x + y for x, y in all_pairs if num_pairs_by_prod[x * y] == 1)

As the sum is enough to know that the integer pair cannot be identified by its product, the sum of the pair cannot be in the above set:

pairs_2 = [(x, y) for x, y in pairs_1 if x + y not in identif_by_prod_pairs_sums]

– P says “Then, I found these numbers.”

This indicates that in `pairs_2`

, the list of pairs restricted by the first two assertions, the correct pair can be identified by its product. Then we can do essentially the same we did in the first step but now *keeping* the pairs identifiable by their products:

num_pairs_by_prod_2 = {} for x, y in pairs_2: if x * y not in num_pairs_by_prod_2: num_pairs_by_prod_2[x * y] = 0 num_pairs_by_prod_2[x * y] += 1 pairs_3 = [(x, y) for x, y in pairs_2 if num_pairs_by_prod_2[x * y] == 1]

– S says “If you could find them, then I also found them.”

This implies that the pair can be identified by its sum from the list restricted by the first three assertions. We can get the final pairs list repeating the previous step, but now searching for a unique sum:

num_pairs_by_sum_3 = {} for x, y in pairs_3: if x + y not in num_pairs_by_sum_3: num_pairs_by_sum_3[x + y] = 0 num_pairs_by_sum_3[x + y] += 1 pairs_4 = [(x, y) for x, y in pairs_3 if num_pairs_by_sum_3[x + y] == 1]

Putting the code together, adding a print statement to get the final list of pairs and running the code we get

[(4, 13)]

matching the results in the literature.

[…] an additional protagonist and a difference sequence of questions and answers, we can try to adapt our previous solution to this new problem. the possible pairs are all combinations of numbers from 0 to 1000, meaning: […]

[…] we have solved a puzzle that claimed to be impossible but was, in fact, just a bit hard But in this post we will see a problem that is really impossible […]

[…] this blog we have solved similar problems before, but this is one that can be easily solved by hand. We only need to be […]