Find the largest palindrome made from the product of two 3-digit numbers.

And the last scratch for now. It is possible to prove that 11 divides a palindromic number. Indeed,

The factor 11 can belong to a - and in this case we step just 1 in b. But if 11 doesn’t divide a, then we can increase b by 11 each time.

This speeds up the previous result around ten times, leaving an asymptotic behavior the same. The memory use is the same O(1).

Let’s look at results:
k = 2 => N = 9009
k = 3 => N = 906609
k = 4 => N = 99000099
k = 5 => N = 9966006699
k = 6 => N = 999000000999
k = 7 => N = 99956644665999
k = 8 => N = 9999000000009999
…
We could improve our algo drastically, if proven that the sought-for palindrome is less or equal (and mirrored). I have the feeling that for even k it is equal. But I don’t know how to prove it. (I calculated for k = 10 and this does not hold, N = 99999834000043899999).

Find the largest palindrome made from the product of two 3-digit numbers.

An author, however, advises a simpler approach. As we are looking for a palindrome a*b, let’s iterate a and b in a top-down direction. After finding some palindrome, impose it as a top boundary for palindromes, that is, iterating in the inner loop for b, we stop when a*b cannot be large than that anymore. If we found a new palindrome, it will replace the boundary. Stop condition is finishing the outer loop in a, i.e. when it drops to 2-digits number (k-1, generally speaking).

Complexity in memory now is just O(1). Performance complexity by my impression is better than in the previous variant. The outer loop has n - n/10 steps, so it cannot be less than O(n). Assuming that a desired palindrome (left half of it, actually) lies close to (which should be proved, strictly speaking), we make no more than operations until find it, and no more than the same afterwards.

This is the worst case, however, and I hope that we find some worse-than-ideal palindrome quick enough. Suppose, we can use the estimate ab origin, i.e. the inequality holds, where f = n - a, g = n - b. Then we can calculate an estimate of operations as area under a curve y = n / x:

So, the actual algo performance is between and .

Find the largest palindrome made from the product of two 3-digit numbers.

Last time we had a straightforward algo with complexity and at least O(n) memory use. Now let’s enhance that. Instead of iterating over multipliers it’s reasonable to iterate over palindromes, starting from the largest. I.e. over sequence 999999, 998899, 997799, and so on.

Remark. The largest product of two 3-digit numbers is 999 * 999 = 998001. So, in principle, we could start from the palindrome 997799. But this saves just 2 iterations.

Having a palindrome m, we factorize it and look at all the subsets of the factorization. Assume, we have one subset already. Let’s name the product of those factors as p. If this number p has k digits (k = 3 for now) and the number m/p has k digits, than we found the palindrome, which is a product of two k-digit numbers.

In Scheme that will be written as:

Here I used a few new util functions:

which make numbers out of their base-k representation.

Complexity now is hard to calculate. The worst case scenario gives quite a bad upper boundary. However, the worst case will never be realized.

Looking at what it gives out (9009, 906609, 99000099, 9966006699, 999000000999, …), I could guess that the required palindrome is found after roughly iterations. So, in total I hope for less than complexity.

The memory use depends on factorizations - we store one whenever a palindrome is taken and lose it when proceed with the next palindrome.

The device is connected to PC and sends out a signal via LPT to a DAC outside. A signal could be rectangular (meander), sinusoidal and ECG-like. (ECG is the electrocardiogram by the way). The ‘heartbeat’ of the latest one can be modulated in time. A subprogram to play out real ECG was included. Some noise modulation was included as well.

That was designed to test cardio-analyzers and similar devices. Small and neat project, as I remember. There was no support for the most pathological arrhythmias, however.

Why not to customize the grid of speed dial in such a way that a user can choose independently x-y size, with discretization 1 column-row? With my usual portrait display mode all the cells are concentrated in the center of a screen.

Something has to be done with shortcuts Ctrl+digit. Nine shortcuts are not enough for 25 cells anymore. I think that could be something like Ctrl+0, [switched to speed dial mode], digit, digit [i.e. up to 99 choices].

A hundred of those tabs should be ok, as for me.

I filed this bugreport them (reference email DSK-258394 at bugs.opera.com).

The same in Haskell:

For imperative languages I’d rather prefer bitwise approach. Here is in C#:

Find the largest prime factor of a composite number.

The problem of integer factorization is one of the most important in the number theory. Last time I implemented a classical algorithm with complexity. The author of the Project Euler website suggests a small improvement — iterating over only odd numbers, but I consider this as too tiny thing to do.

As regards the problem of factorization, I’d rather look in special literature, what are the known approaches (and I’ll do that later, let’s switch to the next one).

Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed four million

The last time we had the straightforward O(n) solution: building a sequence, filtering out even values and adding them. We can improve a bit, noticing that actually, every third member of the Fibonacci sequence is even. We don’t check then for evennes, but just jump over three components each time. This version 2 (I don’t publish it here) should be several times faster, but still is O(n) in performance.

We can also express a member of the Fibonacci sequence via the third and sixth members from behind: and compute those values as the values of a new sequence: . This version 3 is essentially the same as the previous one and again, I don’t publish it here.

The drastic improvement is obtained using the expression (I’ve added it and a proof to the wikipedia article, but they immediately reverted my changes as “unsourced” — this is pathetic). Now the sum is obtained just computing one Fibonacci member, and this can be done with O(log n).

Indeed, we can compute a Fibonacci member exponentiating the appropriate matrix, and this exponentiation, just like usual one, can be done with O(log n). I prefer this solution over using the golden ratio exponentiation formula (again logarithmic complexity), because only integer-operations are involved. So, this is the version 4 of the solution.

I quickly outlined a class for 2D matrices and operations with it:

The solution is O(1) in memory and O(log n) in performance - of course, where n denotes the index of a number in the Fibonacci sequence. And we have been questioned about the cutset, where members of a sequence are less than certain number. Then an additional function (closest-fibonacci-index) comes in handy (see the wiki for explanation):

The final touch is asking ourselves about complexity of the (log) function. Well, it can be computed fast enough not to spoil complexity of the algo’s main part.

Find the sum of all the multiples of 3 or 5 below 1000.

Let us generalize again to a finite set of factors.

There is a formula for the power of finite sets

Here is a measure (i.e. it commutes with the union sign) and can be replaced with — power of a set sign or, if we are in the natural numbers space, with the sum of elements sign, as in our case. is not a multiindex, but a subset of the natural numbers cut from to .

Now by we denote the set of all the multiples of factors , less than certain number N, where i varies from 1 to n (each is respectively the set of multiples of a factor ). We use the above-mentioned formula to compute the measure of the union via measures of all and measures of all finite intersections of them.

Suppose, we have a number , prime or not, and the set of all it’s multiples (they include only numbers less than N). Power of this set is of course (div operation) and the sum of its members can be calculated by the well-known formula for the sum of an arithmetic progression.

As regards all the intersections, it is understandable that we ought to calculate the least common multiple (LCM) of taken factors, and the set-intersection of their multiples will be just a set of its multiples. However, the current version of the solution assumes that we take primes as factors, then the LCM of them is just their product. When I calculate proper LCM in Problem 5 (up to now there is a bruteforce version), I will switch the temp version to it.

Let’s see the solution. New util functions:

The function that calculates subsets of a set:

Important thing about this function is that it returns the empty set as the first element and the full set as the last element of a result list, all other subsets are in between. The number of subsets of a finite subset is just , so the complexity is — it would be better visible with an imperative-iterative version of this function (I’m not posting it here). As regards memory, the function generates all the subsets as lists which in whole contain elements (strange, this neat formula isn’t on Wikipedia yet, I should add it there), that is the memory load is . This is a not-so-good idea to load everything into memory, as we can rewrite this function (and the function that is down here in the post) iteratively with O(n) memory complexity — taking advantage of combinadics, but for now I am satisfied enough with this version.

Using the formula above the solution now as simple as

With that (cdr) I cut off the empty subset, whose measure is zero (otherwise the (sum-of-one) function has to be a bit more complex).

Let’s be careful with notation: n here is actually not the same n, as in the (subsets) function, but the number up there, the maximum of our multiples-sets. The performance complexity depends on k and N, but we are interested only in complexity, depending on N. Let’s assume that k is small comparing to N, which should be the usual case. Then the complexity is roughly speaking O(1), doesn’t depend on N, as we wanted (I remind that in the previous version we had O(N) complexity).

The final touches are the regression tests:

The last commented one breaks, of course, as 15 is not prime - the LCM algo should be updated still.