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Integer TRS Innermost pair #487529016
details
property
value
status
complete
benchmark
sqrt_new.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n140.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.356871128082 seconds
cpu usage
0.323104653
max memory
9601024.0
stage attributes
key
value
output-size
1783
starexec-result
MAYBE
output
/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = condAcc#(false, x, y) -> sqrtAcc#(x, y + 1) sqrtAcc#(I2, I3) -> condAcc#(I3 * I3 >= I2 || I3 < 0, I2, I3) sqrt#(I4) -> sqrtAcc#(I4, 0) R = condAcc(false, x, y) -> sqrtAcc(x, y + 1) condAcc(true, I0, I1) -> I1 sqrtAcc(I2, I3) -> condAcc(I3 * I3 >= I2 || I3 < 0, I2, I3) sqrt(I4) -> sqrtAcc(I4, 0) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = condAcc#(false, x, y) -> sqrtAcc#(x, y + 1) sqrtAcc#(I2, I3) -> condAcc#(I3 * I3 >= I2 || I3 < 0, I2, I3) sqrt#(I4) -> sqrtAcc#(I4, 0) sqrtAcc#(I2, I3) -> sqrtAcc#(I2, I3 + 1) [not(I3 * I3 >= I2 || I3 < 0)] R = condAcc(false, x, y) -> sqrtAcc(x, y + 1) condAcc(true, I0, I1) -> I1 sqrtAcc(I2, I3) -> condAcc(I3 * I3 >= I2 || I3 < 0, I2, I3) sqrt(I4) -> sqrtAcc(I4, 0) The dependency graph for this problem is: 0 -> 3, 1 1 -> 2 -> 3, 1 3 -> 3, 1 Where: 0) condAcc#(false, x, y) -> sqrtAcc#(x, y + 1) 1) sqrtAcc#(I2, I3) -> condAcc#(I3 * I3 >= I2 || I3 < 0, I2, I3) 2) sqrt#(I4) -> sqrtAcc#(I4, 0) 3) sqrtAcc#(I2, I3) -> sqrtAcc#(I2, I3 + 1) [not(I3 * I3 >= I2 || I3 < 0)] We have the following SCCs. { 3 } DP problem for innermost termination. P = sqrtAcc#(I2, I3) -> sqrtAcc#(I2, I3 + 1) [not(I3 * I3 >= I2 || I3 < 0)] R = condAcc(false, x, y) -> sqrtAcc(x, y + 1) condAcc(true, I0, I1) -> I1 sqrtAcc(I2, I3) -> condAcc(I3 * I3 >= I2 || I3 < 0, I2, I3) sqrt(I4) -> sqrtAcc(I4, 0)
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