details

property	value
status	complete
benchmark	collatz.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n137.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.752504 seconds
cpu usage	0.763061
user time	0.432247
system time	0.330814
max virtual memory	124564.0
max residence set size	8596.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = cond2#(false, x) -> f#(3 * x + 1) cond2#(true, I0) -> f#(I0 / 2) cond1#(true, I2) -> cond2#(I2 % 2 = 0, I2) f#(I3) -> cond1#(I3 > 1, I3) R = cond2(false, x) -> f(3 * x + 1) cond2(true, I0) -> f(I0 / 2) cond1(false, I1) -> I1 cond1(true, I2) -> cond2(I2 % 2 = 0, I2) f(I3) -> cond1(I3 > 1, I3) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = cond2#(false, x) -> f#(3 * x + 1) cond2#(true, I0) -> f#(I0 / 2) cond1#(true, I2) -> cond2#(I2 % 2 = 0, I2) f#(I3) -> cond1#(I3 > 1, I3) f#(I3) -> cond2#(I3 % 2 = 0, I3) [I3 > 1] f#(I3) -> f#(I3 / 2) [I3 > 1, I3 % 2 = 0] f#(I3) -> f#(3 * I3 + 1) [I3 > 1, not(I3 % 2 = 0)] cond1#(true, I2) -> f#(3 * I2 + 1) [not(I2 % 2 = 0)] cond1#(true, I2) -> f#(I2 / 2) [I2 % 2 = 0] R = cond2(false, x) -> f(3 * x + 1) cond2(true, I0) -> f(I0 / 2) cond1(false, I1) -> I1 cond1(true, I2) -> cond2(I2 % 2 = 0, I2) f(I3) -> cond1(I3 > 1, I3) The dependency graph for this problem is: 0 -> 6, 5, 4, 3 1 -> 6, 5, 4, 3 2 -> 3 -> 4 -> 5 -> 6, 5, 3, 4 6 -> 3, 4, 5 7 -> 3, 4, 5 8 -> 3, 4, 5, 6 Where: 0) cond2#(false, x) -> f#(3 * x + 1) 1) cond2#(true, I0) -> f#(I0 / 2) 2) cond1#(true, I2) -> cond2#(I2 % 2 = 0, I2) 3) f#(I3) -> cond1#(I3 > 1, I3) 4) f#(I3) -> cond2#(I3 % 2 = 0, I3) [I3 > 1] 5) f#(I3) -> f#(I3 / 2) [I3 > 1, I3 % 2 = 0] 6) f#(I3) -> f#(3 * I3 + 1) [I3 > 1, not(I3 % 2 = 0)] 7) cond1#(true, I2) -> f#(3 * I2 + 1) [not(I2 % 2 = 0)] 8) cond1#(true, I2) -> f#(I2 / 2) [I2 % 2 = 0] We have the following SCCs. { 5, 6 } DP problem for innermost termination. P = f#(I3) -> f#(I3 / 2) [I3 > 1, I3 % 2 = 0] f#(I3) -> f#(3 * I3 + 1) [I3 > 1, not(I3 % 2 = 0)] R = cond2(false, x) -> f(3 * x + 1) cond2(true, I0) -> f(I0 / 2) cond1(false, I1) -> I1 cond1(true, I2) -> cond2(I2 % 2 = 0, I2) f(I3) -> cond1(I3 > 1, I3) popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to Integer TRS Innermost