I set up Doug’s example (actually both of them) and I don’t which is the greater insight: to create a second matrix or to iterate. Neat stuff.

I did something wrong…it takes me 4(!) loops to do the first, and interestingly, 5 to do the second. That turns out to be when the lower right is stable. Maybe Doug knows why the difference. It takes 11 or 12 tries before everything is stable.

I coded up the second method. The seeding Doug mentioned is important. When I did it badly, it took over a minute to calculate and 1218 loops. With the seeds, 3/10′s of a second, and those 5 loops. I’ll post it when I get the time. Needs some cleanup.

…mrt

-Josh

Thank you. It truly never occurred to me that I wanted to be in loop like that. My strategy was to weight the cell I was in so heavily that it would never again be a minimum, and never again visited.

I didn’t succeed. Many times

…mrt

One simplification is that the Min function ignores blank cells, so you can in fact use the same formula all over, and don’t need to adjust it for the edges and corners, or enter big numbers all around the boundaries.

My method has one improvement in that I initially use a specified cell, with non-circular references, to seed the values used in the circular references, rather than using the circular reference from the first iteration. This converges much more rapidly (i.e. 3 iterations, rather than hundreds).

Yep. It’s those circular references that I can’t get my mind around. Thanks for showing an approach. It certainly ought to be doable in a spreadsheet!

…mrt

The procedure is:
1. Open the data as a worksheet (re-name matrix.txt to matrix.csv is a quick way)
2. Insert a blank column to the left, and blank row above the data, so the data range is B2:CC81
3. Set up the worksheet to allow iterative calculation, with 1 iteration, and set calculation to manual.
4. In cell A83 enter 0.
5. Set up an 80×80 table (say in B85:CC164) below the matrix containing the minimum sum to reach any cell:
in cell B85 enter +B2
In cell C86: =+C3+IF($A$83,MIN(B86,C85,D86,C87),B86)
Copy C86 to C86:CB163
In cell C85: =+C2+IF($A$83,MIN(B85,D85,C86),B85) Copy to C85:CB85
In cell CC85: =+CC2+IF($A$83,MIN(CB85,CD85),CB85)
In cell B86: =+B3+IF($A$83,MIN(B85,C86,B87),B85) Copy to B86:B163
In cell B164: =+B81+IF($A$83,MIN(B163,C164),B163)
In cell C164: =+C81+IF($A$83,MIN(B164,C163,D164),B164) Copy to C164:CB164
In cell CC86: =+CC3+IF($A$83,MIN(CB86,CC85,CC87),CB86) Copy to CC86:CC163
In cell CC164: =+CC81+IF($A$83,MIN(CB164,CC163),CB164)
6: Press F9, then enter 1 in cell A83
7: Press F9 (calc) three times.
8: The answer is in cell CC164

An alternative would be to enter a large number (say 1,000,000) in the columns and rows around the range B85:CC164, then use the same formula (as in C86) everywhere except B85.

My VBA essentially used the same technique, and solved near instantaneously.

I have done #81. I’ll post it this weekend, and give you a chance to chime in there. My approach was basically the same as I used for #67 to do #81, swapping min for max, but moving it on to #83 eludes me at the moment.

So thanks, and stay tuned!

…mrt (who was just waiting for fzz to write a sieve for him )

As for Problem 83, I’d be happy to offer as much or as little help as you like. A couple of questions: What approach have you taken so far? Did you solve 81 and 82? How? I solved all three problems with basically the same algorithm.

Josh

No trouble. Reading the problem wrong is an occupational hazard for Eulerites (Eulerians?). More than once I’ve come back to a problem when stuck only to see I was trying to check in the wrong part of the process as the answer. That differs from the times when I read the question right and just flat have a wrong answer

And stuck right now is my usual position. Anybody done #83 please speak up! Being able to go backwards puts me in a infinite loop of minimum cells. I’ve got a conceptual gap. Euler said it’s harder, and I attest that it is.

…mrt

Still, an approach based on the Sieve of Eratasthones (if you have the RAM) is a lot faster than your approach.