## 884 Reputation

16 years, 95 days

@Kitonum Genius.

## You are correct...

@Axel Vogt Zipping the file and its associated images does the trick. Happy Festive Season!

@acer thankyou, that works!

May I bother you with a supplementary Q?

What I want is code (seq or loop) which automatically assigns

Tab[A1] := Record(numwins=0,numlost=0,wins=table([]),losses=table([])):
Tab[B1]:=Record(numwins=0,numlost=0,wins=table([]),losses=table([])):
Tab[A2]:=Record(numwins=0,numlost=0,wins=table([]),losses=table([])):
Tab[B2]:=Record(numwins=0,numlost=0,wins=table([]),losses=table([])):
where A1..B2 are hyperlinked as above.

I know my questions are considered by some to be trivial or even foolish, but I do appreciate your help and patience.

Happy Holidays!

## Same Same...

@acer I repeated the process, this time adding the image files to my email. same problem.

I exported another sheet,  just normal worksheet with &+- input (sans table), and sent it to myself. Got the same problem....

garbled output.

## Ah yes .....

@tomleslie thankyou Sir. Yes, I forgot to change V back to q[n]....

but how might the constraints be printed one line -one constraint as in the pdf:

y1=1

x12+y2=1

x13+x23+y3=1

x14+x24+x34+y4=1

This doesn't solve the problem, but it might assist error tracking.

I changed instances of x[cat(i,",",j)] to index(x,i,j) and z[cat(k,",",i)] to  index(z,k,i).

"In the above formulation qk = 1 if bin k is chosen. This means that a bin is started with item k in the left bottom position. yi = 1 if item i initializes level i. xij = 1 if item j goes to the level initialized by item i. zki = 1 if item i initializes a level in bin k. "

I suppose the statement in the paper would be difficult to program.

From pp4:" z13 = z27 = z2,10 = 1 means that item 3 initializes level 2 in bin 1 while items 7 and 10 initialize two levels in the second bin which is initialized by item 3. "

@acer  As you have alluded, this relates to the project you've been helping me with. I'm trying to hyperlink instances of players names programmatically in the final output table. 4 players =4 different urls. Its proving difficult. Everytime I run it, I get errors......

## Can you...

@Adam Ledger upload the mma and mw worksheets. Or PM me with those attachments.

## Within Table?...

@acer Thanks Mr Saccharum

This is a 3 row table:

with(DocumentTools):with(DocumentTools:-Layout):

T := Table(interior = none, exterior = all, Row("Minesoda SL"), Row("7/11/2017"),Row("Summary Report")):InsertContent(Worksheet(T)):

If I wanted to hyperlink the date. I tried this but I didn't print the date at all, let alone hyperlink it. Is this what you mean by "You cannot get a hyperlink in regular printed output"?

T := Table(interior = none, exterior = all, Row("Minesoda SL"), Row(InsertContent(Worksheet(Group(Input(Textfield(

## subtle difference...

the output is Tab[A1]=Record(......), but what I want is Tab[A1] :=Record(....)

a programmatic approach might work... But i'm trying to keep it in functional style in view of the recent work you did for me.

## picture to text...

@Kitonum Do you have a program to convert screendumps or pictures of math input into maple 1D input?

I hope you don't retype everything from a picture ???

In future the OP should use the big green arrow to upload his worksheet.

## A couple of supplementary questions.......

I wonder if i could impose further:

1/

At the very end there is

#Calculate change in means:

How do I sort this in descending values of mu?

B1: 1168 +- 74, change +5

A1: 1007 +- 42, change +0

........

2/ interum changes:

infolevel[RC]:= 1: #reports

RC: B1 Winner = 1166 +- 78; Change=3; Loser = 1004 +- 45; Change=-3
RC: B1 Winner = 1007 +- 42; Change=3; Loser = 816 +- 44; Change=0

Is it possible to output the winner and loser at each stage like this:
RC: Winner [B1]  = 1166 +- 78; Change=3; Loser [A1] = 1004 +- 45; Change=-3
RC: Winner [A1] = 1007 +- 42; Change=3; Loser [B2] = 816 +- 44; Change=0

## Outstanding work!...

@Carl Love Thanks very much.

I wondered myself how much they charged for the consultation?. The problem is to find such work in the first place. The first barrier is requiring someone at the farm to realise its a scheduling problem and the need for a specialist such as yourself.

In terms of speed, Evolver took less than 5 seconds to get to the minimum for both the 7 block and the 5 block.

The advantage of such a platform is it suits someone with my coding skills (ie very little needed). The disadvantage is that it doesn't always find the minimum. Take the last problem in the document. The number of partitions of it into 3 blocks with the sizes as even as possible is PartCount([2\$5, 4])=945945. (Higher than the 7 block, but lower than 5 block), but Evolver can't seem to get lower than 1205.9 even when I left it going for 45 mins and many trials. (Your procedure gets 1107.1). Now there may be a trick to it, tweaking crossover/mutation rate (it uses genetic algorithms), but I am no expert in it. There may be a forum? This is my worksheet anyways TRP5.xls

As an aside, does Maple do genetic algorithms and neural networks (AI) ?

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