posted                   

code:

---

1
2
2
4
6
8
10
14
20
26
34
46
62
78
102
134
176
226
302
408
528
678
904
1182
1540
2012
2606
3410
4462
5808
7586
9898
12884
16774
21890
28528
37150
48410
63138
82350

---

code:

---

0
0.693147181
0.693147181
1.386294361
1.791759469
2.079441542
2.302585093
2.63905733
2.995732274
3.258096538
3.526360525
3.828641396
4.127134385
4.356708827
4.624972813
4.8978398
5.170483995
5.420534999
5.710427017
6.011267174
6.269096284
6.519147288
6.80682936
7.074963198
7.339537695
7.606884531
7.865571758
8.13446757
8.403352375
8.666991557
8.934059722
9.200087996
9.46374151
9.727585348
9.993785191
10.25864134
10.52271905
10.78746168
11.05307809
11.31873374

---

If you discard the first about 10 itterations I think you could come up with an equation that pretty closley models this, which I would do if I weren't taking a mid-term in under an hour.

Hobbes

Posts: 10602 | Registered: Oct 1999  |  IP: Logged | 

posted                   

code:

---

Length = 35.726*e^(0.2676*X)

---

Where 'Length' is the length of the pattern and 'X' is the pattern number ('1' is X = 1, '11' is X = 2, '2111' is X = 3 and so on).

Hobbes

[ March 31, 2004, 08:59 AM: Message edited by: Hobbes ]

Posts: 10602 | Registered: Oct 1999  |  IP: Logged | 

posted                   

code:

---

1	0	0
0.5	0.5	0
0.75	0.25	0
0.666667	0.333333	0
0.5	0.333333	0.166667
0.5	0.375	0.125
0.6	0.2	0.2
0.571429	0.214286	0.214286
0.6	0.2	0.2
0.5	0.307692	0.192308
0.529412	0.294118	0.176471
0.521739	0.304348	0.173913
0.532258	0.290323	0.177419
0.5	0.307692	0.192308
0.509804	0.27451	0.215686
0.5	0.30597	0.19403
0.5	0.318182	0.181818
0.5	0.30531	0.19469
0.513245	0.301325	0.18543
0.517157	0.301471	0.181373
0.5	0.310606	0.189394
0.488201	0.325959	0.185841
0.503319	0.316372	0.18031
0.50423	0.309645	0.186125
0.494805	0.320779	0.184416
0.497018	0.316103	0.186879
0.494244	0.319263	0.186493
0.495015	0.321114	0.183871
0.497983	0.318915	0.183102
0.496556	0.31801	0.185434
0.494859	0.321645	0.183496
0.496565	0.31865	0.184785
0.493946	0.323347	0.182707
0.494813	0.3199	0.185287
0.495432	0.320923	0.183646
0.495583	0.319651	0.184766
0.494429	0.320927	0.184644
0.495063	0.320078	0.184859
0.495011	0.320869	0.18412
0.495701	0.319611	0.184687

---

You'll notice all three columns comfortably settle on one number to atleast two significant figures. So the equation produces a string of numbers that increases size exponentialy, but at a very steady exponential rate (a low degree of error from the above posted exponential equation) and results in a stable distribution of numbers.

I'm impressed that a recursive equation (as all patterns are) as... arbitrary as this yields such stability.

A programmers note: this is indeed a recursive problem, and it set itself up very well for a recursive solution. However, the limitor on it was more in terms of memory than CPU time, and trying to put that all on the stack for each function would've eaten up memory even faster than the itterative solution I setteled on. It is possible to make this a memory efficent program with dynamic memory allocation, deleting the old memory in a different function call than it was declared, but I began to worry about bad pointers, and besides, itterative was safer and easier. I'm thinking of going back and making it a program that uses very little memory by writting to disk right away and increasing the step quickly, but that would take a while and I still have 3 more midterms in the next 48 hours so we'll see.

Hobbes

[ March 31, 2004, 10:17 AM: Message edited by: Hobbes ]

Posts: 10602 | Registered: Oct 1999  |  IP: Logged | 

posted                   

code:

---

#!/usr/bin/env python
from time import sleep

basefilename = 'stupid'

startseqfilename = basefilename + '1'

readseqfilename = startseqfilename
if __name__ == '__main__':
	while(1):
		writeseqfilename = basefilename + str(int(readseqfilename[len(basefilename):]) + 1)
		readseqfile = file(readseqfilename, 'r')
		writeseqfile = file(writeseqfilename, 'w')
		xxx = readseqfile.read(1)
		currentcount = 0
		currentnum = xxx
		while xxx != '' and xxx != '\n':
			if currentnum == xxx:
				currentcount += 1
			else:
				writeseqfile.write(str(currentcount) + currentnum)
				currentcount = 1
				currentnum = xxx
			xxx = readseqfile.read(1)
		if currentcount != 0:
			writeseqfile.write(str(currentcount) + currentnum)
		readseqfile.close()
		writeseqfile.close()
		readseqfilename = writeseqfilename

---

code:

---

#!/usr/bin/env python
from seq import basefilename
testfile = file(basefilename + '1')
basenum = testfile.readline()
baselen = len(basenum)
if basenum[baselen - 1] == "\n":
	basenum = basenum[:baselen - 1]
resultfilename = 'results' + str(basenum)

howmany = 20
results = file(resultfilename, 'w')
for ii in range(1, howmany + 1):
	seqfilename = basefilename + str(ii)
	seqfile = file(seqfilename, 'rU')
	seqdata = seqfile.readline()
	if seqdata[len(seqdata) - 1] != "\n":
		results.write(seqdata)
		results.write("\n")
	else:
		results.write(seqdata)
	results.flush()
	seqfile.close()
results.close()

---

Posts: 15770 | Registered: Dec 2001  |  IP: Logged |